Math 50 T9a-Right Triangle Trigonometry Review Page MTH 50 TOPIC 9 RIGHT TRINGLE TRIGONOMETRY 9a. Right Triangle Definitions of the Trigonometric Functions 9a. Practice Problems 9b. 5 5 90 and 0 60 90 Triangles Review 9b. Practice Problems 9c. Determining Lengths of Sides 9c. Practice Problems 9d. Reference ngles and Trigonometric Functions 9d. Practice Problems
Math 50 T9a-Right Triangle Trigonometry Review Page 9a. Right Triangle Definitions of the Trigonometric Functions In this section, we present the definitions of the trigonometric functions based on a right triangle. n alternate but equivalent method of introducing the trigonometric functions is based on the unit circle (Review Topic ). Consider some acute angle,. Form a right triangle, where C = π =90. (Remember that the sum of the interior angles in any triangle is 80 or π radians.) B C Fig. 9a. We have the following definitions. sin = opposite hypotenuse = BC C cos = adjacent hypotenuse = C B tan = opposite adjacent = sin cos = BC C cot = adjacent opposite = tan = cos sin = C BC sec = hypotenuse adjacent csc = hypotenuse opposite = cos = B C = sin = B BC
Math 50 T9a-Right Triangle Trigonometry Review Page In a similiar fashion, we could start with the other acute angle in the right triangle, angle B, and define sin B = opposite hypotenuse = C B, BC cos B = B, etc. Notice that opposite and adjacent are relative to the angle in question but that the definitions do not change. lso remember that since C is a right angle, ( π ) B =90 = radians.
Math 50 T9a-Right Triangle Trigonometry Review Page PRCTICE PROBLEMS for Topic 9a-Right Triangle Definitions of the Trigonometric Functions B 5 C 9a.. Find exact values for the following. a) sin = b) cos = c) tan = d) sinb = e) cos B = f) cotb = g) sec = h) sin +cos = i) csc B = ( π ) j) sin = 9a.. Find exact values for all remaining trig ratios given that sec =.
Math 50 T9a-Right Triangle Definitions of the Trigonometric Functions nswers Page 5 NSWERS to PRCTICE PROBLEMS (Topic 9a-Right Triangle Definitions of the Trigonometric Functions) 9a.. a) sin = 5 ; b) cos = 5 ; c) tan = ; d) sinb = 5 ; e) cos B = 5 ; f) cotb = ; g) sec = 5 ; h) sin +cos =; i) csc B = 5 ; j) sin ( π ) = 5 ; 9a.. sec = implies we have the following picture. The Pythagorean Theorem implies that the missing side must equal 7 6 9= 7. Thus, sin =, cos =, 7 tan =, cot =, and csc =. 7 7
Math 50 T9b-5 5 90 and 0 60 90 Triangles Review Page 6 9b. 5 5 90 and 0 60 90 Triangles Review The two types of triangles mentioned in the title often appear in calculus. For the acute angles in these triangles, we will find the exact values of the associated trigonometric functions by studying the relationships between the functions, the lengths of the sides, and the angles. We begin with Fig. 9b.. B Given: B = π C = π rad = 5 C = π rad = 90. Fig. 9b. Since the sum of the interior angles is π or 80, B also = π radians. This means that the triangle is isosceles, which implies C = BC. By the Pythagorean Theorem, (C) +(BC) =, (BC) =, BC = = = C. Thus we have π Fig. 9b. B π C This implies that sin π =, cos π =, tan π ==cotπ, etc.
Math 50 T9b-5 5 90 and 0 60 90 Triangles Review Page 7 Next consider a 0-60 -90 triangle. B 60 0 Fig. 9b. C Given: =0 = π 6 rad B =60 = π rad C =90 = π rad B = We now form Fig. 9b., which comes from flipping the triangle in Fig. 9b. about the line C. B 60 0 0 C 60 D Fig. 9b. Evidently, BD is an equilateral triangle since each vertex angle is 60. This means BD =andbc = CD =, since C is perpendicular to BD. (*See remark below.) So we have 0 Fig. 9b.5 60 B C
Math 50 T9b-5 5 90 and 0 60 90 Triangles Review Page 8 The Pythagorean Theorem enables us to write C = ( ). Specifically, = +(C), or C =. Using right triangle definitions (Review Topic 9a), we see that sin 0 =sin π 6 =, cos 0 =cos π 6 =, sin 60 =sin π =, cos 60 =cos π =, tan 0 = =, tan 60 = =, etc. *Remark: There is a popular formula that says the side opposite the 0 angle in a right triangle is one half the hypotenuse. Now we see why this is correct. Based on Figures 9b. and 9b.5, we can fill in the following table. θ π 6 π π sin θ cos θ tan θ Table 9b.6 PRCTICE PROBLEMS for Topic 9b-5 5 90 and 0 60 90 Triangles 9b.. Study Figs. 9b. and 9b.5 and be able to reconstruct them by memory. Try to understand how the various quantities are derived. 9b. Fill in Table 9b.6.
Math 50 T9b-5 5 90 and 0 60 90 Triangles Review Page 9 9b. Construct an isosceles right triangle with legs of length. Find the sine, cosine, and tangent of π and compare these values to those found in Fig. 9b.. What conclusions can be drawn?
Math 50 T9b 5 5 90 and 0 60 90 Triangles nswers Page 0 NSWERS to PRCTICE PROBLEMS (Topic 9b 5 5 90 and 0 60 90 Triangles nswers) 9b. θ π 6 π π sin θ cos θ tan θ nswer 9b. 9b. The values of the trigonometric functions in Fig. 9b. will be the same for any isoceles right triangle.
Math 50 T9c Determining Lengths of Sides Review Page 9c. Determining Lengths of Sides Right triangle trigonometry can also give the lengths of the sides of a right triangle, if you know at least one of the acute angles and one side of the triangle. For example, in the figure below, find BC and C. B 0 Fig. 9c. C Since sin 0 =sin π 6 = (see Review Topic 9b), we can write BC =sin =, or BC =6. Then C can be obtained by suing the cosine or the tangent function (or the Pythagorean Theorem). For example, = tan = BC C. Thus = 6 C,orC =6. Note that since =0 (and C =90 ), angle B =60. Then we could apply the same reasoning to angle B and obtain the lengths of the respective sides. Let s do another example where you have to use a calculator. Consider Fig. 9c., where angle B =50 and C =. Find the lengths of B and BC. 50 B Fig. 9c. C
Math 50 T9c Determining Lengths of Sides Review Page Using a calculator (set in degree mode!) sin 50 =.766. Thus, Next,.766 = sin 50 =sinb = B, or B = =5. (approximately)..766 tan B = tan 50 =.9 = BC, or BC =.9 =.6. In calculus, sometimes the following problem is encountered. If sin θ = x, and θ is in the first quadrant, find cos θ and tan θ. To solve this problem, we first note that x = x. This allows us to create the following picture, since sin θ = opposite hypotenuse. θ x Fig. 9c. By the Pythagorean Theorem, we can solve for the missing side (call it a), at least algebraically. That is, c = a + b, or = a + x, or a = x. So we now have the following. θ x x Fig. 9c. This means cos θ = x, and tan θ = x x.
Math 50 T9c Determining Lengths of Sides Review Page PRCTICE PROBLEMS for Topic 9c-Determining Lengths of Sides 9c.. B Fig. 9c.5 C Suppose =5 = π radians. If BC =, find C and B. 9c.. B C Fig. 9c.6 Suppose B = 8 and angle = radian. Find C and BC. 9c.. If cos θ = x and θ is in the first quadrant, find sin θ and tan θ.
Math 50 T9c Determining Lengths of Sides nswers Page NSWERS to PRCTICE PROBLEMS (Topic 9c-Determining Lengths of Sides) 9c.. tan π =. lso, tan π = BC BC. Thus, = C =. Note that C C since the triangle is isosceles, elementary geometry also implies that C =. Next sin π = = BC B = B. Thus B =. (Reminder: We can also find B by using the Pythagorean Theorem.) 9c.. Using a calculator set in radian mode, we find that sin() =.8. This means BC =.8, or BC =6.7. Next, cos() =.5. Since 8 cos = C B = C, we have C =8(.5) =.. 8 9c.. Given cos θ = x we can draw the following picture. θ x Fig. 9c.7 The Pythagorean Theorem implies the missing side has length x. Then, sin θ = x, and tan θ = x. x
Math 50 T9d Reference ngles and the Trigonometric Functions Review Page 5 9d. Reference ngles and the Trigonometric Functions The definitions of the trigonometric functions in Review Topic 9a were based on having an acute angle (less than 90 ). What happens if > 90 = π radians? For example, Let be the angle pictured in Fig. 9d.. Now, what is meant by sin, cos, etc.? Fig. 9d. To answer this question, we consider the reference angle of, which is the acute angle between the terminal side of and the x axis. In this case, the reference angle is (π ) as in Fig. 9d.. (For the rest of our discussion, the reference angle of will also be denoted by ref.) ref ref Fig. 9d. Fig. 9d. We next form a right triangle between the terminal side of and the x axis, as indicated in Fig. 9d.. Then we define sin = sin(ref angle). Note that since (ref ) is an acute angle with its own right triangle, we can again use definitions involving an opposite side, adjacent side, etc., just as we did for Fig. 9a. in Review Topic 9a.
Math 50 T9d-Reference ngles and the Trigonometric Functions Reviews Page 6 For example, consider Fig. 9d. below. 5 ref Fig. 9d. Then we define sin = sin(reference angle)= 5,, cos = 5, tan = =, etc. Observe that the adjacent side of the reference angle is ( ), since x is negative. lso note that since we have a right triangle, the numbers,, and 5 satisfy the Pythagorean Theorem. The situation is similar if the terminal side of lies in the third or fourth quadrant, as in Fig. 9d.5. ref 5 Fig. 9d.5 Fig. 9d.6 We form a right triangle with the reference angle as indicated in Fig. 9d.6, and we suppose the lengths of the sides are as shown. Then sin = sin(reference angle) = 5 5, cos =, tan =, etc.
Math 50 T9d Reference ngles and the Trigonometric Functions Review Page 7 PRCTICE PROBLEMS for Topic 9d-Reference ngles and the Trigonometric Functions 9d.. Consider angle as in Fig. 9d.7, and suppose its reference angle forms a right triangle with the dimensions shown in Fig. 9d.8. Find sin, cos, tan. ref 5 Fig. 9d.7 Fig. 9d.8 Evidently, based on Figures 9.d, 9d.6, and 9d.8, we can conclude that sin θ > 0forθ in the first and second quadrants and sin θ < 0forθ in the third and fourth quadrants. Similarly, cos θ is positive in the first and fourth quadrants and negative in the second and third quadrants, while tan θ is positive in the first and third quadrants and negative in the second and fourth quadrants. The table below summarizes these observations. st Q. nd Q. rd Q. th Q. sin θ + + cos θ + + tan θ + + Table 9d.9 - Q means quadrant.
Math 50 T9d-Reference ngles and the Trigonometric Functions Reviews Page 8 PRCTICE PROBLEMS for Topic 9d-Continued 9d.. Determine whether the following expressions are positive or negative. a) tan 7π 6, b) sec π, c) sin π 6, d) sin5π 6, e) cos 5π, f) tan θ, if π <θ<π. 9d.. Name the quadrant(s) where θ must terminate, given the following information. a) sin θ<0 b) sin θ>0 and tan θ<0. Finally, we can use the idea of reference angles to extend what we know about 5-5 -90 and 0-60 -90 triangles from the first quadrant to the other quadrants. For example, let = π. Let B, C, and D be the 6 positive angles in the second, third, and fourth quadrants respectively that have reference angle π 6. moment s reflection yields that B = 5π 6,C= 7π 6, and D = π 6. = π 6 ref B B = 5π 6 ref C C = 7π 6 D = π 6 ref D Fig. 9d.0. ll reference angles above = π 6.
Math 50 T9d Reference ngles and the Trigonometric Functions Review Page 9 s shown in Review Topic 9b, sin π 6 =. This leads to Fig. 9d.. θ = π 6 θ Fig. 9d. Figs. 9d.0 and 9d. immediately yield the following information. θ = 5π 6 θ = 7π 6 θ = π 6 Fig. 9d.. Fig. 9d. enables us to create the table below. θ sin θ cos θ 5π 6 7π 6 π 6 tan θ Table 9d.
Math 50 T9d-Reference ngles and the Trigonometric Functions Reviews Page 0 We remark that similar reasoning applies for negative angles. For example, if θ = π, it has the same reference angle as the drawing on the right in 6 ( Fig. 9d. when θ = π ) (. Thus, sin π ) = ( 6 6, cos π ) = 6, etc. s an additional example, θ = π would have the same reference angle as θ = 5π. ( nother observation is that sin π ) ( π ) ( = sin and cos π ) = ( 6 6 6 π ) cos. We will see in Review Topic that this observation is true in 6 general. That is, for any x, sin( x) = sin x (sin x is an odd function), cos( x) = cos x (cos x is an even function), and tan( x) = tan x (tan x is an odd function). The concept of even and odd functions is explained in Review Topic. PRCTICE PROBLEMS for Topic 9d-Continued 9d.. 9d.5. What are the positive angles in the second, third, and fourth quadrants that have π as their reference angle? What are the positive angles in the second, third, and fourth quadrants that have π as their reference angle?
Math 50 T9d Reference ngles and the Trigonometric Functions Review Page 9d.6. rguing as we did in Fig. 9d.0, 9d., and 9d., complete the tables below. θ π π 5π sin θ cos θ tan θ Table 9d. θ π 5π 7π sin θ cos θ tan θ Table 9d.5 9d.7. Find the exact value for each expression. ( a) cos π ) ( b) tan π ) c) sin ( π )
Math 50 T9d-Reference ngles and the Trigonometric Functions nswers Page NSWERS to PRCTICE PROBLEMS (Topic 9d-Reference ngles and the Trigonometric Functions) 9d.. sin = 5 5, cos =, tan = = 5. 9d.. a) positive b) negative, since sec θ = cos θ. c) negative d) positive e) negative f) negative 9d.. a) rd quadrant or th quadrant b) nd quadrant 9d.. π, π, 5π 9d.5. π, 5π, 7π 9d.6. θ π π 5π sin θ cos θ tan θ Table 9d.5
Math 50 T9d-Reference ngles and the Trigonometric Functions nswers Page θ π 5π 7π sin θ cos θ tan θ Table 9d.5 9d.7. a), b) c). NOTE: The solution for part a) can be derived in two ways. First, if we use the concept of reference angle, then θ = π has reference angle π. From Table 9b.6, cos π =. Since π is in the ( fourth quadrant where the x coordinate is positive, we have cos π ) =. s a second approach, since cos x is an even function, (i.e., cos( x) = ( cos x, we can simply say cos π ) ( π ) =cos =. Similar comments apply for answers b) and c). Beginning of Topic Skills ssessment