MPE Review Section II: Trigonometry Review similar triangles, right triangles, and the definition of the sine, cosine and tangent functions of angles of a right triangle In particular, recall that the corresponding sides of similar triangles are proportional Moreover, two triangles are similar if two angles of one are equal to two angles of the other Don't overlook the enormously important Pythagorean Theorem for right triangles, (a + b c ) You will need to use a scientific calculator for the trigonometry section of the Colorado State University Mathematics Placement Examination In particular, review how and when to use radian or degree measure for angles, and how to evaluate trigonometric and inverse trigonometric functions ANGLES The angle θ shown in the Figure T is in standard position; ie, its initial side is the positive x-axis and its terminal side is a ray from the origin The angle is positive if the rotation is counter-clockwise, and negative if the rotation is clockwise Example: Estimate the degree and radian measures of the angle θ in Figure T Solution: Angle θ is formed by rotating a ray through / revolutions counterclockwise from the positive x-axis One full revolution counterclockwise forms an angle of 60 or radians Therefore, the measures of angle θ are ( 60 ) 0 6 or ( ) What is the approximate degree measure of the angle in Figure T? Figure T Figure T What is the approximate degree and radian measures of the angle in Figure T? Figure T - -
Since one counterclockwise revolution forms an angle of 60 or radians, the equations for converting between degrees and radians are and radians radians 60 80 radian 60 80 degrees degrees What is the radian measure of 08 angle? 7 What is the degree measure of an angle of radians? 8 What is the degree measure of an angle of radians to the nearest tenth of a degree? 7 ARC LENGTH If an arc of length s on a circle of radius r subtends a central angle of radian measure θ, then s rθ Example: Consider a circle with radius What is the arc length intercepted by a angle? Solution: First convert to radians: And so with θ we have s rθ s 6 Determine the arc length intercepted by a 78 angle in a circle of radius 7 Determine the arc length intercepted by AREA OF A SECTOR 9 radians in a circle of radius Consider the sector determined by the angle θ, in the circle with radius, r The arc length of a 80 is Recall that the area of the circle is given by A r θ Since the sector is the fractional part of the circle, the area of the sector then is given by θ r θ A r Figure T - - Figure T
Example: Consider a circle with radius What is the area of a sector determined by a angle? Solution: First convert to radians: And so with θ we have ( ) A 8 Find the area of a sector determined by a 96 angle in a circle of radius 6 9 Find the area of a sector determined by THE TRIGONOMETRIC FUCTIONS radians in a circle of radius For an angle θ in standard position, let (x, y) be a point on its terminal side By the Pythagorean Theorem r + x y The six trigonometric functions of θ are defined as y r sin θ, csc θ, r y x cos θ, r y tan θ and x r sec θ, x x cot θ y Figure T6 For an acute angle θ of a right triangle, as shown in Figure T7, the trigonometric functions can also be defined in terms of the side opposite angle θ, the side adjacent to angle θ, and the hypotenuse Thus, side oppositeθ hypotenuse, cscθ, hypotenuse side oppositeθ side adjacent toθ hypotenuse, secθ, hypotenuse side adjacent toθ side oppositeθ side adjacent toθ tan θ and cotθ side adjacent toθ side oppositeθ Figure T7 - -
Notice that cot θ, sec θ, and csc θ are respectively the reciprocals of tan θ, cos θ and sin θ, and that the fundamental identity sin θ + cos θ is just another way of expressing the Pythagorean Theorem By examining the various cases, we can see that the sine function is positive for angles whose terminal side is in the first or second quadrant and negative for angles in quadrants III and IV Similarly, the cosine is positive in quadrants I and IV and negative in quadrants II and III; while the tangent is positive in quadrants I and III and negative in quadrants II and IV Figure T8 Example: The angle φ is a quadrant II angle and tan φ What is the value of cosφ? 8 Solution: Draw a figure so that φ has its terminal side in quadrant II and so that tan φ 8 8 Then φ Hence, r x + y () + ( 8) 89 8 8 89 cosφ 89 89 Figure T9 Example: Find all angles θ between 0 and such that tan θ -79 Solution: We need to determine the quadrants of the two angles θ andθ between 0 and such that tan θ -79 Since tan θ -79 is negative and the tangent is negative in quadrants II and IV, θ is in quadrant II and θ is in quadrant IV Next we need to find the reference angle for θ andθ tan θ 79 79 θ 70 Make a sketch of the angles θ andθ in the correct quadrants with reference angle θ 70 as in Figure T0 Determine θ andθ using Figure T0 as a guide: θ θ 70 96 and θ θ 70 06 Figure T0 - -
0 Given sin θ and cot θ is negative, what is the value of 9 a) cos θ (exactly)? b) 6cos θ - cot θ (rounded to two decimal places)? In Problems -, use a calculator to approximate the answer to five decimal places Given that angle φ is a quadrant II angle and tan φ, find 6 sin φ + sec φ 8 9 Angle θ is a quadrant III angle and csc θ What is the value of tan θ - sec θ? The point (0,-) is on the terminal side of angle θ in standard position What is the value of tan θ + csc θ? What is the value of sin (86 ) sec (6 ) + 8 tan (-96 )? 0 What is the value of sin csc() + 7 cos? 7 6 Angles α and β are between 0 and 60 with cos α -0906 and tan β -0877 Which of the following is a possible (approximate) value for α + β? A) B) 9 C) 09 D) 6 E) 669 SOLVING RIGHT TRIANGLES Given a right triangle and some data about it, we say we've "solved" the triangle when we know the measures of its three sides and three angles In the special case of a right triangle ABC, with C 90 (see Figure T), we are able to solve the triangle if we know either ) the measure of two sides, or ) the measures of one side and one additional angle Figure T Example: Given triangle ABC with C 90 and a 6, b, solve the triangle Solution: By the Pythagorean Theorem c (6) + () 77 Hence, the hypotenuse c 77 7 (approximately) Now we can calculate 6 sin A 0 77 so that A 7 Then, B 90-7 6 NOTE: All calculator work is rounded, so results are necessarily approximate - - Figure T
Example: Given a right triangle ABC with C 90 and a 7, B, solve the triangle Solution: Make a sketch (even if the proportions are wrong), see Figure T Since 7 b cos B and tan B, c 7 we may find 7 7 c 69 cos B cos and b 7 tan B 7 tan 09 Finally, A 90-6 Figure T 7 What is the approximate perimeter of the triangle ABC where a 0, A and C 90? 8 In triangle ABC, b 69, c 88 and C 90 What is the approximate value of A - B? 9 In triangle ABC, a 6, b 88 and C 90 Find c, A and B SOLVING OBLIQUE TRIANGLES A triangle which does not have a right (90 ) angle is called an oblique triangle An oblique triangle either has three acute (less than 90 ) angles, or it has two acute angles and one obtuse (greater than 90 ) angle Figures T and T show the possibilities These figures also show the standard labeling The angles (or vertices) are denoted by capital letters and the sides opposite the angles are denoted by the same letters in lower cases As with all triangles, the sum of the angles of an oblique triangle ABC is 80, A + B + C 80 Figure T Figure T The equations relating sides and angles of right triangles are not true for oblique triangles The basic relationships among sides and angles of an oblique triangle are given by the Law of Sines and the Law of Cosines When three parts (including a side) of an oblique triangle are given, the Law of Sines or the Law of Cosines can be used to solve the triangle There are four possible cases: - 6 -
Case I: Case II: Case III: One side and two angles are given Two sides and one opposite angle are given Two sides and the included angle are given Case IV: Three sides are given The Law of Sines is used in Cases I and II The Law of Cosines is used in Cases III and IV LAW OF SINES For any triangle ABC, Example: Case I: One side and two angles given Solve ΔBQK where a, Q 0 and K Solution: In this case, the angle B is immediately found as a sin A B 80 ( Q + K) 80 We can solve for q by the Laws of Sines: so q sin0 sin q sin 0 668 sin We can solve for k by the Law of Sines: so b sin B c sin C k sin sin k sin 008 sin 0 Find side b in ABC if a, A 80 and B 6 Solve PDQ where q 7, P 7 and Q Figure T6 Figure T7 Case II: (Given two sides and one opposite angle) may also be solved by the Law of Sines provided a solution exists Case II is called the ambiguous case because there may be two solutions, or just one solution, or no solution at all Use your review references to study the various possibilities The figures on the next page show some of the possibilities for solutions (or no solution) given sides a and c, and angle A - 7 -
Figure T8 Two solutions Figure T9 No solutions Figure T0 One solution Figure T One solution Example: Solve ABC where a 0, c and A Solution: We attempt to solve for angle C by the Law of Sines: 0 sin C sin so sin sin C 0 0 There is no angle C for which sin C 0 Hence, there is no triangle having these sides and angles Example: Solve ABC where a 8, c 0 and A Solution: By the Law of Sines, 0 8 0 sin so sin C 000 (rounded) sin C sin 8 Hence C 90 Thus, in this case there is exactly one solution To complete the solution, we find B 80 ( A + C) 80 ( ) 687 Finally, solve for b by the Law of Sines: b 0 0sin 687 so b 60 sin 687 sin 90 sin 90-8 -
Example: Solve ABC where a, c 7 and A 0 Solution: By the Law of Sines, 7 sin C 7 sin0 so sin C 090 sin 0 There are two angles C between 0 and 80 for which sin C 090 They are C 7007 and C 099 Since and A + C 0 + 7007 007 < 80 A + C 0 + 099 99 < 80, C and C each lead to a solution of the triangle There are two triangles with given sides and angle We must find them both When C C 7007, B 80 - (7007 + 0 ) 99 According to the Law of Sines, then b sin99 so b 8 sin 99 sin 0 sin0 The first solution to this triangle is a, b 8, c 7, A 0, B 99, C 7007 When C C 099, B 80 - (099 + 0 ) 007 Again by the Law of Sines, b sin007 so b 986 sin 007 sin 0 sin0 The second solution to this triangle is a, b 986, c 7, A 0, B 007, C 099 Example: Solve ABC where a, c 6 and A 70 Solution: By the Law of Sines, 6 6sin 70 so sin C 07 sin C sin 70 There are two angles C between 0 and 80 for which sin C 07 They are C 7 and C Since A + C 70 + 7 97 < 80, C can be an angle of a triangle which has the given sides and angle Since A + C 70 + > 80, - 9 -
C cannot be an angle of a triangle with the given sides and angle This triangle has only one solution It comes from C 7 When C C 7, B 80 - (70 + 7 ) 8 By the Law of Sines, then, b sin 8 sin 70 so sin 8 b 77 sin 70 The only triangle with sides and angle as given has a, b 77, c 6, A 70, B 8, C 7 Solve ABC where a, c and A Solve ABC where a 89, c 000 and A 6 Solve ABC where a 9, c 76 and A 986 In oblique triangle ABC, a 07, c and C 98 Which one of the following might be angle A to the nearest tenth of a degree? A) B) 88 C) 7 D) 6 E) none of these LAW OF COSINES For any triangle ABC labeled in the customary way (as in Figure T), a b + c - bc cos A Figure T Notice that the Law of Cosines can also be stated as and b a + c - ac cos B c a + b - ab cos C When one angle of a triangle is 90, the Law of Cosines reduces to the Pythagorean Theorem Example: Case III: Given two sides and the included angle Solve ABC where a 7, b 9 and C 7 Solution: By the Law of Cosines Thus Hence c 7 + 9-7 9 cos 7 c 9 + 8-898 068 c 66-0 -
Now we may solve for angle A by the Law of Cosines a 9 + (66) 7 b + c bc cos A so cos A ()(9)(66) Then A 06 Now we can find angle B as B 80 - (06 + 7 ) 8 The only triangle with given sides and angle has a 7, A 06, b 9, B 8, - - c 66, C 7 0666 Example: Case IV: Given three sides Solve ABC where a 6, b and c Solution: We begin by solving for angle A (We could begin with any angle) a b + c - bc cos A so 6 + - cos A Then + 6 cos A 00666 Hence, A 98 Next we solve for angle B by the Law of Cosines b a + c - ac cos B so 6 + - 6 cos B Then 6 + cos B 08666 6 Hence, B 99 Now we can easily find angle C as C 80 - (98 + 99 ) 6 Thus the complete solution is a 6, A 98, b, B 99, c, C 6 Example: Case IV: Given three sides Solve ABC where a, b and c Solution: When we apply the Law of Cosines, we find and + - cos A cos A + 9 6 > There is no angle A with cosine greater than, so there is no triangle having these three sides An attempt at cos B or cos C will also yield numbers outside the interval [-, ] One can also see that there is no solution by noticing that there are two sides (a and c ) whose sum is less than the third side, + <
6 Solve ABC with b, c and A 0 7 Solve ABC with a, b and c 7 8 Find angle R in PQR where p, q 0 and r 9 9 Solve PAS with p 7, a and s 0 Solve ABC with a 7, b 7 and c 7 NARRATIVE PROBLEMS In the following examples and problems it is important that you read the problem carefully, draw a reasonable sketch of the situation, identify and label the known and the unknown quantities, write an equation relating the known and unknown quantities, solve the equation and 6 examine the reasonableness of your answer as it relates to the sketch Example: The Washington Monument casts a shadow 79 feet long when the angle of elevation of the sun is How tall is the monument? (Assume the ground is level, though you may know differently) Solution: From the sketch as shown in Figure T, x tan 79 Hence x 79 tan 6 feet Example: Two men, 00 feet apart on level ground, observe a hot-air balloon between them The respective angles of elevation are measured as 6 and 78 What is the altitude of the balloon above the ground level? h Solution: Since sin C, h a sin 7 8 a We need to find side a By the Law of Sines, a sin A b sin B Since B 80 - (6 + 78 ) 699, we get Thus, a 00 sin 6 sin 699 Figure T Figure T - -
00sin 6 a 7 sin 699 Hence, h a sin 78 (7) sin 78 9 feet A surveyor determines the following information about a triangular shaped piece of property One side has length 79 feet A second side has length 8 feet The angle opposite the 8 foot side measures 8 How long is the third side of the property? A merchant ship at its dock in New York was observed to subtend an angle of 6 from the window of an office 000 feet from the bow and 000 feet from the stern of the ship How long is the ship? A vertical tower is braced by two cables which are anchored to the ground at the same point At this point, the angle between the cables is The first cable is 0 feet long and extends to the top of the tower The second cable extends to a point 0 feet below the top of the tower How long is the second cable? As an airplane is taking off, its path of ascent makes a 60 angle with the runway How many feet does the plane rise while it travels 00 feet in the air? (Assume the path of the ascent is a straight line) GRAPHING TRIGONOMETRIC FUNCTIONS Values of the trigonometric functions at 0, 0,, 60 and 90 should be memorized The following table gives the values degrees radians sine cosine tangent cosecant secant cotangent 0 0 0 0 undefined undefined 0 6 60 90 0 undefined undefined 0 Example: Using this table (not a calculator) find sin sec Solution: From the table we have sin and sec Hence, cos sin sec - -
Evaluate csc 0 - cot 0 6 Evaluate cos sin tan 7 Evaluate sec + tan 8 Evaluate sec cot 9 Which of the following is the exact numerical value of? cot A) B) 6 C) D) E) GRAPHS OF THE SINE AND COSINE FUNCTIONS The graph of the function y sin θ is shown in Figure T The horizontal (or θ ) axis is in radians, so the graph crosses the horizontal axis at 0, ±, ±, ±, etc The maximum value of the sine function is, which occurs at and ± n for each integer n The minimum value of the sine function is - The section of the graph from θ 0 to θ is one complete cycle, ie, the graph repeats itself from to, from to 6, etc Similarly, it repeats from - to 0, from - to -, etc Figure T The sine function The graph of the cosine function is shown in Figure T6 It, too, repeats in cycles of and has values between - and If the cosine graph is displaced units to the right, it coincides with sine graph, ie, cos(θ - ) sin θ Thus, both sine and cosine complete their cycle in units Figure T6 The cosine function - -
The function y sin θ has the same cycle length as y sin θ, namely but y sin θ rises to maximum value of (when θ, for instance) and falls to a minimum of - (at, for instance) Its graph is shown below in Figure T7 Figure T7 The maximum height above the horizontal axis of a graph that oscillates equally above and below the x-axis is called the amplitude Thus, y sin x has amplitude, y cos x has amplitude, and y - sin x has amplitude Now consider the function y sin θ Its amplitude is, but because θ runs from 0 to while θ is running from 0 to, this function completes one cycle in units At a glance its graph looks exactly like the sine function The only difference is that it cycles faster (twice as fast) as the ordinary sine function Its graph is shown below in Figure T8 Figure T8 The period of this function is, which is the length of one cycle Now we combine these ideas A function of the form y A sin Bθ or y A cos Bθ has Amplitude A and Period B 0 Find the amplitude and period of the function y sin θ Find the amplitude and period of the function y - cos θ θ Find the amplitude and period of the function y cos Find the amplitude and period of the function shown in Figure T9 Find the equation for the function Figure T9 6 - -
Find the amplitude and period of the function shown in Figure T0 Find the equation for the function Figure T0 The graphs of the tangent and cotangent functions are shown in Figure T and Figure T Figure T Graph of y tan θ Figure T Graph of y cot θ You should be able to construct these graphs Note that both tangent and cotangent functions have period rather than The tangent function has vertical asymptotes at odd multiples of, while the cotangent function has asymptotes at multiples of (that's the same as even multiples of ) Between two consecutive asymptotes, the tangent function is increasing while the cotangent function is decreasing The graph of the secant function appears in Figure T - 6 -
Figure T Graph of y sec θ Figure T Graph of y csc θ The cosine function is also plotted so that you can see the reciprocal relationship sec θ between these functions Both the secant and cosecant have period For instance, one complete section of the graph of the secant lies between θ and θ The secant function has vertical asymptotes at odd multiples of (where the cosine function is 0) The graph of the cosecant function is shown in Figure T It is just the secant graph displaced units to the right You should be able to construct the graphs of the secant and cosecant functions The period of y csc θ is 6 The period of y tan θ is 7 The period of y cos θ is 8 The period of y cot θ is 9 The period of y sin θ is 0 The period of y sec θ is Sketch the graph of the function y cot θ by making a table of values Use angles θ in the interval 0 θ which are multiples of Figure Ta - 7 -
Sketch the graph of the function y csc θ by making a table of values Use angles θ in the interval 0 θ which are multiples of Figure Tb Sketch the graph of the function y sec θ by making a table of values Use angles θ in the interval θ which are multiples of Figure Tc INVERSE TRIGONOMETRIC FUNCTIONS A portion of the graph of y sin x is shown in Figure T6 Over the interval x to x, the sine function takes on each of its values exactly once Consequently, for each y between - and + there is exactly one number x between and for which y sin x In this way the portion of the graph of y sin x between x and x determines a function which associates a unique number between and with each number between - and This function is called the inverse sine function and is denoted Figure T6 y sin x - 8 -
y sin - x Figure T7 shows the graph of y sin - x The domain of the inverse sine function is - x and its range is y In this setting, when y is considered an angle, the unit of measure is radians, not degrees Figure T7 y sin - x The inverse cosine function y cos - x and the inverse tangent function y tan - x are defined in similar ways Figure T8 shows the graph of y cos - x The domain of y cos - x is - x and its range is 0 y Figure T8 y cos - x Figure T9 shows the graph of y tan - x The domain of the inverse tangent function is < x < and its range is < y < Figure T9 y tan - x In the functions y sin - x, y cos - x and y tan - x, think of y as an angle expressed in radians When using a scientific calculator to work with these functions, set the calculator to radian mode Example: Find sin Solution: Set sin y By the definition of the inverse sine function, sin y and y The angle between and whose sine is is the familar special angle Thus, - 9 -
sin This function evaluation involved a familiar special angle,, and so should be performed by inspection ( as was done in the solution above) You can verify that the end result is correct by calculating either Example: Find tan ( ) - sin 07070678 or sin 078986 - Solution: Set y tan ( ) - By the definition of the inverse tangent function, tan y and < y < Since tan y is negative, y must lie in the interval < y < 0 between and 0 whose tanget is is, so tan ( ) - The angle Since this evaluation involved only a familiar special angle, it should be performed by inspection Of course, the result can be verified by calculation In Problems -6, find the value of the inverse function without using a calculator sin - tan - 6 cos - (-) 7 sin - () In Problems 7-9, find approximate values of the inverse trigonometric functions using a scientific calculator Round off the final answer to (four) decimal places 8 cos - (-07) 9 sin - (0990) 60 tan - (-000) Many applications involve compositions of trigonometric and inverse trigonometric functions such as sin(sin - x), sin - (sin x) and sin(cos - x) The simplest of these are of the form sin(sin - x), cos(cos - x) and tan(tan - x) Consider the first of these If y sin - x, then x sin y and hence, sin(sin - x) sin y x Thus, as long as x is in the domain of the inverse sine function, - 0 -
sin(sin - x) x Similarly, if cos - x is defined, then cos(cos - x) x If tan - x is defined, then tan(tan - x) x 6 Evaluate cos (cos - (-09)) 6 Evaluate sin (sin - 0) 6 Evaluate tan (tan - (09)) 6 Evaluate sin (sin - (-9)) Simplifying compositions such as sin - (sin x), cos - (cos x) and tan - (tan x) requires more care Example: Evaluate tan - (tan 07)) - Solution: Since 07 lies in the interval < x <, tan (tan07) 07 8 Example: Evaluate cos - cos 7 Solution: The range of the inverse cosine function is the interval from 0 to Since 8 8, > is not in the range of the inverse cosine function Consequently, 7 7 8 8 cos - cos 7 7 Our problem is to find an angle θ between 0 and (in the range of cos - ) such that 8 cos 7 8 A quadrant II angle having the same reference angle as will meet this requirement 7 8 The reference angle for is The quadrant II angle with reference angle is 7 7 7 6 θ 7 7 8 6 6 Since cos cos and is in the range of cos - 7 7 7-8 - 6 6 cos cos cos cos 7 7 7 Verify this value using a scientific calculator 6 Evaluate tan (tan - 870) 66 Evaluate cos (cos - 7) 67 Evaluate tan - (tan ) 68 Evaluate tan - (tan(-067)) 69 Evaluate sin - (sin 00) 70 Evaluate cos - cos 7 Evaluate cos - (cos (-)) 7 Evaluate tan - (tan ) - -
7 Evaluate tan - tan 7 Evaluate sin - sin - - 7 Evaluate sin - sin 6 76 Evaluate cos - cos Finally, we evaluate composition such as sin(cos - x) and tan(sin - x) of a trigonometric function with some other inverse trigonometric function These compositions cannot be evaluated directly from the definitions of the inverse trigonometric functions - Example: Evaluate sin cos Solution: Set y cos - We must find sin y Since cos y > 0, y is a quadrant I angle as shown in Figure T0 By the definition of the cosine a a function cos y r a + b where (a, b) is any point on the terminal side of angle y To find sin y, we choose a point (a, b) so that a a + b Figure T0 The easiest choice is a, so a + b 6 + b 6 + b b ± since the terminal side of the angle is in quadrant I, b + The point (a, b) (, ) and, by the definition of the sine function, - sin cos sin y - Example: Simplify cos tan Solution: Set tan - Hence, tan and y y < y < Since tan y is negative, y lies in the interval < y < 0 and y is a quadrant IV angle Sketch angle y in standard position in quadrant IV as shown in Figure T Figure T
Choose a point (a, b) on the terminal side of y so that b tan y a It is easiest to choose this point so a and b - The distance between the origin and the point (a, b) is r ( ) + 9 - Hence, cos y and cos tan 9 9 - - 77 Evaluate cos sin 78 Evaluate sin tan - 79 Evaluate tan cos 80 Evaluate - cos( tan ) - 8 Evaluate sin cos 8 Evaluate cos( sin - ( ) ) - 8 Evaluate sin( tan ) - 8 Evaluate tan( cos ) - 8 8 Evaluate cos tan 86 Evaluate - tan sin 7-87 What is the exact value ofsin tan? A) B) 9 C) D) 9 69 BASIC TRIGONOMETRIC IDENTITIES E) 69 An identity is an equation that is true for all values of the variables for which each member of the equation is defined An identity that relates trigonometric functions is called a trigonometric identity The reciprocal identities csc θ, secθ, and cotθ tanθ and the quotient identities tan θ and cotθ are easily derived from the definition of the trigonometric functions of an angle θ in terms of a point (x, y) on the terminal side of the angle The variables x, y and r in the definitions of the trigonometric functions are related by the equation - -
so x + y r x y + r r Therefore, by the definitions of the sine and cosine functions, cos θ + sin θ Similarly, + tan θ sec θ and cot θ + csc θ These identities are called the Pythagorean identities The reciprocal, quotient and Pythagorean identities are used frequently Memorize them or learn to construct them quickly by reasoning from the definitions as in the discussion above Memorize the basic identities above Then, without referring to the statements of the basic identities, fill in the blanks to form one of the reciprocal, quotient or Pythagorean identities, or one of their alternate forms 88 tan θ + 89 90 sin θ + cos θ 9 9 9 - - 9 csc θ 9 cos θ tanθ 96 csc θ cot θ 97 tan θ cotθ 98 99 tan θ sec θ cotθ 00 cos θ 0 sin θ cotθ 0 cscθ 0 0 cos θ tanθ 0 Given the quadrant of an angle and the value of one of its trigonometric functions, we can use the preceding identities to find the values of the other trigonometric functions Example: Find cot θ if tan θ - Solution: Since cotθ, we have cotθ tanθ - secθ cscθ
Example: Find if andθ is a quadrant II angle Solution: Since cos θ + sin θ, cos θ + or cos θ Then,, where the - sign is chosen because the cosine is negative in quadrant II Example: Find cotθ if andθ is in quadrant III Solution: Since none of the basic identities involves just cot θ and cos θ, we first find sin θ and then find cotθ as Now sin θ + 8 So sin θ and The negative sign is chosen because θ is in 9 quadrant III Finally, cot θ Example: Find if tanθ andθ is in quadrant IV Solution: First, cotθ Second, from the Pythagorean identity cot θ + csc θ, we get ( ) + csc θ csc θ Since θ is a quadrant IV angle, cscθ Finally, cscθ Use only the reciprocal identities, quotient identities and Pythagorean identities to solve Problems 06 06 Find secθ if tanθ andθ is a quadrant IV angle 07 Find csc θ if 7 08 Find if secθ 09 Find cotθ if cscθ andθ is a quadrant I angle 0 Find tanθ if andθ is a quadrant II angle - -
Find cotθ if and 7 Find if secθ andθ is a quadrant IV angle Find secθ if cscθ andθ is a quadrant II angle Any trigonometric expression can be written entirely in terms of the sine and cosine functions by using the reciprocal and quotient identities Example: tanθ secθ ( ) Example: csc cos θ + sin θ θ + sec θ + sin θ cos θ sin θ cos θ sin θ cos θ Write each expression in terms of the sine and cosine functions and then simplify the expression sec θ + tan θ csc θ 6 cotθ Here are some suggestions for verifying trigonometric identities successfully secθ tanθ + cotθ Become very familiar with the reciprocal, quotient and Pythagorean identities in all their forms Begin with the more complicated side of the identity and, using the basic identities, try to reduce it to the simpler side If necessary, convert all functions to the sine and cosine and then proceed to simplify each side Example: Verify the identity sec θ - sec θ - tan θ Solution: Begin with the more complicated right-hand side and factor it sec θ tan θ ( sec θ + tan θ )( sec θ tan θ ) ( sec θ + tan θ ) sec θ + tan sec θ + ( sec θ ) sec θ Example: Verify the identity tan θ + cot θ csc θ sec θ Solution: Since no terms are squared, the Pythagorean identities are not helpful Begin by writing the left side in terms of the sine and cosine Verify each of the following identities sin θ + cos θ tanθ + cotθ + cscθ secθ 7 cos θ - sin θ cos θ - 8 cot θ + cot θ + csc θ θ - 6 -
9 ( + tan θ ) cos θ 0 sin θ sin θ - sin θ cos θ csc θ ( cos θ + sin θ ) cot θ + sec θ + + + 0 sec θ ( ) 7 9 cotθ + tanθ + 6 cos θ sin θ ( tan θ + cot θ ) secθ tanθ 8 secθ + tanθ + tan θ secθ cscθ + cotθ cscθ cotθ 0 cos 6 θ - sin θ + sin θ - sin 6 θ NEGATIVE-ANGLE, SUM, DIFFERENCE AND COFUNCTION IDENTITIES If θ is an angle in standard position and (x, y) is a point on its terminal side, then (x, -y) is a point on the terminal side of -θ Also, (x, y) and (x, -y) are the same distance r from the origin Consequently, sin cos ( θ ) y, r x ( θ ) and tan( θ ) tanθ The trigonometric identities r y x sin(-θ ) -sin θ, cos(-θ ) cos θ and tan(-θ ) -tan θ are called the negative-angle identities They should be memorized The difference identity for the cosine function says that for any two angles θ and φ cos (θ - φ) cosφ + sinφ The derivation of this identity involves the distance formula and the Law of Cosines and is somewhat intricate Read it in your favorite trigonometry reference To help you remember the difference identity for the cosine, remember that when θ φ this identity reduces to the Pythagorean identity cos θ + sin θ By replacing φ by -φ in the difference identity for the cosine and then using the negative angle identities, we obtain the sum identity for the cosine: cos (θ + φ) cosφ - sinφ By considering quadrant I angles (especially angles smaller than / radians) one can see that if θ is an angle in standard position and (x, y) is a point on its terminal side, then (y, x) is a point on the terminal side of / - θ Also, (x, y) and (y, x) are the same distance from the origin From the definition of the - 7 -
trigonometric functions of an angle in terms of a point on the terminal side of the angle, it follows that for any angle θ sin θ tan θ cotθ cos θ cot θ tanθ These relationships, called the cofunction identities, are also useful in degree form sin(90 - θ ) tan(90 - θ ) cotθ cos(90 - θ ) cot(90 - θ ) tanθ Memorize the negative-angle identities, the difference and sum identities for the cosine function, and the cofunction identities Then, without referring to the statements of these identities, fill in the missing members of the following identities cos(θ + φ ) -tanθ sin (90 - θ ) cos (-θ ) cosφ + sinφ 6 cot θ 7 sin (-θ ) 8 cosφ - sinφ 9-0 cos θ tan (90 - θ ) cos (θ - φ ) tan (-θ ) We can also use these identities to find values of the trigonometric functions for certain special angles Example: Find cos (0 ) Solution: cos( 0 ) sin( 90 0 ) sin( 60 ) sin 60 Example: Find cos θ +, if and 0 0 Solution: cos θ + cos sin 0 0 0 0 0-8 -
In Problems -6, evaluate each expression using the negative-angle identities tan 6 cos (90 ) + sin (-80 ) 6 sin tan In Problems 7-9, evaluate each expression using the cofunction identities 7 cos 8 cot 9 tan (90 - ) + cos (90-80 ) In Problems 0-, use the sum and difference identities for the cosine function to evaluate each expression 0 cos cos cos 0 7 7 cos cos sin sin cos( 0 θ ) if and cos θ + if and By replacing θ with - θ in the difference and sum identities for the cosine and then applying the cofunction identities, we obtain the sum and difference identities for the sine function: sin (θ + φ ) cosφ + sinφ and sin (θ - φ ) cosφ - sinφ The sum and difference identities for the tangent function are developed from the sum and difference identities for the sine and cosine and the quotient identities These identities state that for any angles θ and φ tanθ + tanφ tan( θ + φ ) tanθ tanφ and tanθ tanφ tan( θ φ ) + tanθ tanφ - 9 -
You should either memorize all of these identities or learn to reconstruct them quickly by reasoning similar to that in the discussion Example: Verify the identity cos (80 - θ ) -cos θ Solution: cos ( 80 θ ) cos80 + sin80 + 0 Example: Verify the identity ( θ + φ ) cos cotθ tanφ cosφ Solution: We apply the identity to the quantity cos (θ + φ) and then simplify the fractions ( θ + φ ) Verify each of the following identities cos cosφ sinφ cosφ sinφ cosφ cosφ cosφ cosφ sinφ cotθ tanφ cosφ cos (θ + ) - 6 cos (θ + φ ) + cos (θ - φ ) cosφ 7 tan (-θ ) - sec θ - 8 cos cos DOUBLE-ANGLE AND HALF-ANGLE IDENTITIES ( θ φ ) cos( θ + φ ) ( θ φ ) + cos( θ + φ ) The double-angle identities are obtained by setting θ φ in the three sum identities: and sin θ sin (θ + θ ) +, cos θ cos (θ + θ ) - cos θ - sin θ tan tanθ + tanθ tanθ tanθ ( θ ) tan( θ + θ ) Thus, the double-angle identities are: and sin θ, cos θ cos θ - sin θ tanθ tan θ tan θ tanθ tanφ The half-angle identities are derived from the double-angle identities for the cosine From the Pythagorean identity cos θ - sin θ, so cos φ cos φ - sin φ - sin φ By solving for sin φ, we obtain - 0 -
cos φ sinφ ± θ Finally, by replacing φ with, we obtain θ sin ±, θ where the sign is determined from the quadrant of By a similar calculation θ + cos ±, where the sign is determined by the quadrant of θ Finally, to obtain a formula for tan θ, use the half-angle identities for the sine and cosine in and get θ sin tan cos θ tan ± ( θ ) ( θ ), + where, again, the sign is determined by the quadrant of You should memorize the double-angle and half-angle identities by learning to reconstruct them quickly from previous identities as in the discussion above 9 sin θ 60 sin θ cos φ - cos θ sin φ 6 - sin θ 6 tan (θ + φ ) 6 θ + ± 6 sin (θ + φ ) 6 cos θ 66 ± + 67 cos θ - 68 sin θ tanθ tanφ 69 + tanθ tanφ 70 sin (θ - φ ) 7 tan θ 7 cos θ - sin θ 7 sin θ cos θ 7 sin θ cos φ + cos θ sin φ - -
7 tanθ tan θ The half-angle, double-angle, and sum and difference identities can be used to evaluate the trigonometric functions for certain special angles The following examples illustrate this use Example: Evaluate sin 7 Solution: Since 7 0 +, we have sin 7 sin (0 + ) sin 0 cos + cos 0 sin + + Example: Evaluate cos 8 Solution: Apply the half-angle identity for the cosine to the angle 8 + cos + cos cos + 8 6 Notice that the + sign is chosen for the radical because 8 is a quadrant I angle and the cosine is positive in quadrant I 8 Example: Find sin θ, if sin θ and θ is a quadrant IV angle 7 Solution: By the double-angle identity, sin θ sin θ cos θ 8 Since sin θ, the missing piece in this puzzle is the value of cos θ Since 7 sin θ + cos θ, we have 8 + cos θ 7 so that 6 cos θ 89 89 and hence, cos θ ± 7 We must choose the + sign because θ is a quadrant IV angle and the cosine is positive in quadrant IV Thus, - -
cos θ 7 and 8 0 sin θ 7 7 89 Use trigonometric identities to evaluate each expression 76 sin sin 6 77 cos 6 - sin 6 78 θ 79 Find sin if andθ is a quadrant I angle 80 Find cos θ if sin θ + φ if, sinφ and both θ and φ are quadrant I 8 Find tan ( θ + φ ) if tanθ and tanφ 8 Find ( ) angles sin cos We may use the recently developed identities to verify more complicated trigonometric identities Example: Verify the identity sin (θ + φ ) sin (θ - φ ) sin θ - sin φ Solution: From the sum and difference formulas for the sine, we have sin ( θ + φ ) sin( θ φ ) ( cosφ + sinφ )( cosφ sinφ ) sin θ cos φ cos θ sin φ sin θ ( sin φ ) ( sin θ ) sin θ sin θ sin φ sin φ + sin θ sin φ sin θ sin φ sec θ Example: Verify the identity sec θ sec θ θ Solution: θ sec sec cos θ cos θ sec θ sec θ In the following problems, verify that each equation is a trigonometric identity cotθ cotφ 8 cot( θ + φ ) cot θ 8 cot θ cotθ + cotφ cotθ 8 ( ) + sin θ θ 86 cos sin sin θ 87 sin θ sin θ (HINT : θ θ + θ ) sin φ θ - -
TRIGONOMETRIC EQUATIONS Most of the trigonometric equations we have seen up to this point have been identities; ie, they were true for all values of the variables Now we wish to consider conditional equations ones which are true for some, but not all, values of the variables To solve a conditional equation means to find all values of the variables for which the equality holds Example: The trigonometric equation sin θ is true for some values of θ and not for others It is not true for θ 6, neither is it true for θ your social security number There are an infinite number of solutions to this equation They are given by θ ± n, where n is an integer Some trigonometric equations don't have any solutions; ie, they are not true for any of the variables Example: Solve - Solution: Rewrite this equation as sin θ + cos θ For every value of θ, sin θ + cos θ < Therefore, there are no values of θ for which sin θ + cos θ Hence, this equation has no solutions There is no general procedure for solving all trigonometric equations and some can be very difficult For example, you might try your hand at the innocent-looking tan θ θ (After you run out of good ideas with identities and algebraic manipulations, you might get out your calculator and try some approximations) We will be concerned only with equations that can be solved by using trigonometric identities to change the form of the equation and applying algebraic methods such as rearranging terms, factoring, squaring, and taking roots The simplest trigonometric equations are sin θ c, cos θ c and tan θ c, where c is a constant Inverse trigonometric functions can be used to find one solution to equations of this form, when the equation has a solution Our problem is to use this information to find all solutions The equations c and c have two solutions in the interval 0 θ < when c <, and one solution when c Since the sine and cosine functions have period, to find all solutions, add and subtract integer multiples of to the solutions in the interval 0 θ < For any number c, the equation tanθ c has one solution in the interval 0 θ < Since the tangent function has period, to find all solutions, add integer multiples of to the solution in the interval 0 θ < An expression (or expressions) which describes all solutions to a trigonometric equation in a simple way is called the general solution to the equation Example: Solve the trigonometric equation cos θ Solution: One solution is cos - θ The second solution in the interval 0 θ < is - -
θ θ The only solutions in the interval 0 θ < are θ and θ angles to represent the general solution: θ + n or θ + n, where n is any integer All of these solutions can be represented in a single expression as θ ± + n, n an integer Example: Find the general solution of the equation tanθ We use these Solution: One solution is θ Tan - 6 The general solution can be represented as θ + n, n an integer 6 However, it is customary to represent the general solution in terms of a solution between 0 and The angle θ is not in this interval, so we use θ + to represent 6 the general solution as θ + n, n an integer 6 In problems 88-9, find the general solution to each trigonometric equation 88 sin θ 89 tan θ - 90 cos θ 0 9 Equations of the form sin Bθ c, cos Bθ c and tan Bθ c, where B and c are constants, are slightly more complicated than the equations just considered Since sin Bθ and cos Bθ have period (positive and negative) integer multiples of, to express the general solution to sin Bθ c and cos Bθ c, add B to the solutions in the interval 0 θ < The B B so the general solution to tan Bθ c is expressed by adding integer function tan Bθ has period B multiples of to the solutions in the interval 0 θ < B B - -
The technique for solving these equations is to transform them into the form of the simpler equations just considered by making a change of variables Example: Find all solutions to sin θ in the interval 0 θ < Solution: Make the change of variables φ θ so the equation becomes sin φ The general solution to the original equation is φ + n, φ + n, n an integer Thus, the general solution to the original equation is θ + n, θ + n, n an integer or θ + n, θ + n, n an integer 8 8 The solutions in the interval 0 θ < are 9 θ,,, 8 8 8 8 In problems 9-9, find all solutions in the interval 0 θ < 9 9 cos θ 9 tan θ 0 θ sin 9 sin θ We now extend our techniques to solve equations that yield to rearranging and factoring Example: Find all solutions of tanθ in Solution: Rewrite tanθ as tanθ - 0 and factor to obtain (tanθ - ) 0 Either 0 or tanθ - 0-6 - 0 θ < We solve these equations separately in the stated interval cos θ 0 gives solutions θ, tanθ 0 (or tanθ ) gives solutions and Thus, the solutions in the interval 0 θ < are θ, and θ,
Example: Find all solutions of sin θ + in Solution: Rewrite the equation as sin θ - + 0 and factor as a quadratic in to obtain ( - )( - ) 0 0 θ < Set each factor equal to 0 and solve the two equations - 0 and - 0 in the interval 0 θ < The first yields solutions and, while the second 6 6 yields θ The solutions in the interval 0 θ < are θ, and θ 6 6 In problems 96-00, find all solutions in the interval - 7-0 θ < 96 sin θ 97 cos θ - 98 cos θ - cos θ 99 + + 00 sin θ sin θ There are two situations in which identities are commonly used to solve trigonometric equations When the equation involves more than one trigonometric function, an identity may be used to rewrite the equation in terms of just one function When the trigonometric equation involves an unknown angle θ and its multiples, we may use the double-angle, half-angle and addition formulas to rewrite the equation in terms of just one angle These techniques are illustrated in the following examples Example: Find all solutions of sec θ + tanθ in 0 θ < Solution: We use the identity sec θ + tan θ to rewrite the equation as + tan θ + tanθ Subtract from both sides and factor to get tanθ (tanθ + ) 0 Thus tanθ 0 or tanθ + 0, so the solutions in θ 0, and θ, 7 Example: Find all solutions of - 0 in the interval 0 θ < are 0 θ < Solution: We use the identity - sin θ to rewrite the equation as - ( - sin θ ) 0 which is now in terms of θ (and not θ ) Solving by factoring,
sin θ + 0 ( + )( ) 0 or sin θ The solutions in the interval 0 θ < are and θ θ, 6 6 Example: Find all solutions of + in the interval 0 θ < Solution: Rewrite the equation as - Square both sides to obtain cos θ - + sin θ Use the identity cos θ - sin θ to obtain - sin θ - + sin θ Now solve by factoring sin θ 0 ( ) 0 sin θ 0 or sin θ The solutions in the interval 0 θ < are θ 0, and θ Finally, the squaring process in the first step may have introduced extraneous roots We check the values θ 0,, in the original equation to see if they really are solutions θ 0 : sin 0 + cos0 0 + θ 0 is a solution θ : sin + cos + 0 θ is a solution θ : sin + cos 0 + ( ) θ is not a solution Thus, the solutions are θ 0, θ In problems 00-06, find all solutions of the given equation in the interval 0 θ < 0 cot θ - 0 0 cos θ - sin θ 0 cos θ sin θ + 0 + cot θ csc θ 06 07 + tanθ secθ - 8 - cos θ
II TRIGONOMETRY ANSWER SECTION ANGLES -0-76, 7 radians radians 70-9 ARC LENGTH 6 76 7 6 AREA OF A SECTOR 8 68 9 07 THE TRIGONOMETRIC FUNCTIONS 77 0 (a) ; 9 (b) 6-76 668 tan θ is undefined -88-7769 6 C) 09 SOLVING RIGHT TRIANGLES 7 070 8-9 9 c 6, A, B 69 SOLVING OBLIQUE TRIANGLES 0 b 0 D 8, d, p 77 ) C 77, B 8, b 67 ) C 06, B 07, b 77 no solution C 8, B 66, b E) none of these (A 679 ) LAW OF COSINES 6 a 899, B 87, C 7 no solution ( + < 7) 8 68 9 P 66, A 90, S 77 ( + 7 ) 0 A B C 60 NARRATIVE PROBLEMS 97 feet,0 feet feet,09 feet GRAPHING TRIGONOMETRIC FUNCTIONS 6 7 7 8 undefined 9 E) - 9 -
GRAPHS OF THE SINE AND COSINE FUNCTIONS 0 amplitude ; period amplitude ; period amplitude ; period 6 amplitude ; period ; cos( θ ) 6 y amplitude ; period ; y sin θ 6 7 8 9 0 see page 6, Figure T see page 7, Figure T see page 7, Figure T INVERSE TRIGONOMETRIC FUNCTIONS 6 6 6 7 undefined ( is not in the domain of Sin - x) 8 76 9 00 60-0000 6-09 6 0 6 09 6 no solution 6 870 66 undefined 67 68-067 69 00 70 7 0 7 0 7 7 6 7 76 77 78 79 80 8 8 0 8 8 undefined 8 7 86 8 87 D) 69 BASIC TRIGONOMETRIC IDENTITIES 88 sec θ 89 cot θ 90 9 sec θ 9 tan θ 9 + cot θ 9 cot θ 9 - sin θ 96 97 98 tan θ 99-00 sin θ 0 cos θ 0 0 cos θ 0 sin θ 0 sin θ 06 7 07 7 08-09 0-7 + cot θ 6 sin θ 7 cos θ - sin θ cos θ - ( - cos θ) cos θ - - 0 -
8 cot θ + cot θ + (cot θ + ) (csc θ) csc θ sin θ + tan θ cos θ + + cos θ θ θ θ 9 ( ) cos cos sin 0 - cos θ ( - cos θ ) sin θ csc θ ( + ) + + θ + ( ) cot sin θ + ( ) ( + ) + sin θ + cos + cotθ + tanθ + sec sin θ cos θ θ ( ) ( ) ( cos θ + sin θ ) 0 θ + sec θ ( ) ( sin θ ) sec θ + sec θ 6 ( tanθ + cotθ ) + sin θ + cos θ secθ tanθ secθ tanθ 7 secθ tanθ secθ + tanθ secθ + tanθ secθ tanθ sec θ tan θ 8 9 cscθ + cotθ ( cscθ + cotθ ) cscθ cotθ csc θ cot θ cscθ cotθ cscθ cotθ cscθ cotθ + + secθ ( + secθ ) tan θ sec θ (secθ + )(secθ ) secθ sec θ cos θ 0 cos 6 θ ( - sin θ ) ( - sin θ )( - sin θ + sin θ ) - sin θ + sin θ - sin 6 θ NEGATIVE-ANGLE, SUM, DIFFERENECE AND CO-FUNCTION IDENTITIES cosφ - sinφ tan (-θ ) cos (θ - φ) 6 tanθ 7-8 cos (θ + φ) 9 sin (-θ ) 0 cotθ cosφ + sinφ -tanθ 0 6 7 8-9 0 + 6 6-0 - -
cos (θ + ) cos - sin - - 0-6 cos (θ + φ) + cos (θ - φ) cosφ - sinφ + cosφ + sinφ cosφ 7 tan (-θ ) - sec θ (tan (-θ )) - sec θ (-tan θ ) - sec θ tan θ - sec θ - 8 cos cos ( θ φ ) cos( θ + φ ) ( θ φ ) + cos( θ + φ ) ( cosφ + sinφ ) ( cosφ sinφ ) ( cosφ + sinφ ) + ( cosφ sinφ ) sinφ tanθ tanφ cosφ DOUBLE-ANGLE AND HALF-ANGLE IDENTITIES 9 60 sin (θ - φ) 6 cos θ 6 tanθ + tanφ tanθ tanφ 6 cos θ 6 cosφ + sinφ 6 cos θ - sin θ cos θ - - sin θ 66 tan θ 67 cos θ 68 ± 69 tan (θ - φ) 70 cosφ - sinφ 7 tanθ tan θ 7 cos θ 7 sin θ 7 sin (θ + φ) 7 tan θ 76 6 77 78 79 80 8 6 8 6 8 7 9 8 cot( θ + φ ) 8 tan tanθ tanφ tanθ tanφ cotθ cotφ cotθ cotφ ( θ + φ ) tanθ + tanφ tanθ + tanφ cotθ cotφ cotθ + cotφ tan θ tan θ cot cot θ tan θ tanθ tanθ cot + sin θ sin θ + + cos θ cot θ θ cotθ θ 8 ( ) 86 sin θ θ θ cos cos sin 87 sin sin( θ + θ ) θ θ cos θ + sin θ ( sin θ ) + ( ) sin sin θ + cos θ θ sin θ + ( sin θ ) - -
TRIGONOMETRIC EQUATIONS 88 θ + n, θ + n, n 0, ±, ±, 89 θ + n, n 0, ±, ±, 90 θ + n, n 0, ±, ±, 9 θ + n, θ + n, n 0, ±, ±, 9 θ,,, 9 θ 0,,,,, 7 9 no solution 9 θ,,,,,,, 6 6 7 96 θ,,, 97 θ,, θ 0 6 6 6 6 7 98 θ,,, 99 θ, θ 0 6 6 6 6 7 00 θ 0,,,, θ, 0 θ,,, 8 8 8 8 7 7 0 θ,,, 0 θ,,,,, 6 6 9 9 9 9 9 9 0 θ 0, 0 θ, 06 θ 0 07 θ 0 6 6 - -