SECTION TWO: TRIGONOMETRIC FUNCTIONS ND THEIR TLES:. TRIGONOMETRIC FUNCTIONS ND THEIR TLES: Let P(,) be an point on the unit circle: Unit Circle: + = P(,) - - - If P is the terminal point that corresponds to an angle, then we can define the following functions:. COSINE: The -coordinate of the point P is called the cosine of the angle and denoted b cos Cosine Function Domain: R, Range: [,] 0,, is in the first quadrant is in the second quadrant cos > 0 P(,) P(,) cos < 0 - - Unit Circle: + = Unit Circle: + =
, 3 3, is in the third quadrant cos < 0 is in the forth quadrant cos > 0 P(,) - - P(,) Unit Circle: + = Unit Circle: + =. SINE: The -coordinate of the point P is called the sine of the angle and denoted b sin Sine Function Domain: R, Range: [,] 0,, is in the second quadrant sin > 0 is in the first quadrant sin > 0 P(,) P(,) - - Unit Circle: + = Unit Circle: + =
, 3 3, is in the third quadrant is in the forth quadrant sin < 0 sin < 0 P(,) - - P(,) Unit Circle: + = Unit Circle: + = SUMMRY: n point P on the unit circle coterminal to an angle has coordinates cos,sin Unit Circle: + = P(cos, sin ) - n immediate result that follows summar: cos sin = Eplanation: Since P is an point on the unit circle, its coordinates must satisf the equation of the unit circle. Therefore we get cos sin = Notation: cos is denoted as cos,and sin is denoted as sin, etc.
3. TNGENT: The -coordinate of the point Q is called the tangent of the angle and denoted b tan (0,) Q (, tan ) P(cos, sin ) (-,0) (,0) (0,-) Unit Circle: + = From the figure, it is understood that the -coordinate of point Q is well-defined 3 ecept when is different from,, etc. Therefore the domain of the tangent function is R { k }, where k is an integer. Since Q can be anwhere on its ais, it takes an value in the interval,. Therefore the range of the tangent function is,, namel all real numbers. Tangent Function Domain: R { k }, k Z, Range:, 0,, is in the first quadrant tan > 0 (0,) Q (, tan ) P(cos, sin ) is in the second quadrant tan < 0 P (0,) (-,0) (,0) (-,0) (,0) (0,-) Unit Circle: + = (0,-) Q Unit Circle: + = (, tan )
, 3 3, is in the third quadrant tan > 0 (0,) Q (, tan ) is in the forth quadrant tan < 0 (0,) (-,0) P (0,-) (,0) Unit Circle: + = (-,0) (0,-) P (,0) Q Unit Circle: + = (, tan ) 4. COTNGENT: The -coordinate of the point R is called the cotangent of the angle and denoted b cot (-,0) (0,) R (cot, ) P(cos, sin ) (,0) (0,-) Unit Circle: + = Once again, from the figure, it is understood that the -coordinate of point R is welldefined ecept when is different from 0,, etc. Therefore the domain of the cotangent function is R {k }, where k is an integer. Since R can be anwhere on its ais, it takes an value in the interval,. Therefore the range of the cotangent function is,, namel all real numbers. Cotangent Function Domain: R {k }, k Z, Range:,
0,, is in the first quadrant cot > 0 (-,0) (0,) R (cot, ) P(cos, sin ) (,0) R is in the second quadrant cot < 0 (cot, ) (-,0) P (0,) (,0) (0,-) Unit Circle: + = (0,-) Unit Circle: + =, 3 3, is in the third quadrant cot > 0 (0,) R (cot, ) R is in the forth quadrant cot < 0 (cot, ) (0,) (-,0) (,0) (-,0) (,0) P (0,-) Unit Circle: + = (0,-) P Unit Circle: + = CONCLUSION: In the unit circle, the -ais ma be called the cosine ais, the -ais ma be called the sine ais. The ais on which point Q travels is called the tangent ais, and the ais on which point R travels is called the cotangent ais. (See figure below)
ais (or sine ais) R (0,) Q cotangent ais P (-,0) θ (,0) ais (or cosine ais) P ( cos, sin ) Q (, tan ) R ( cot, ) (0,-) tangent ais 5. SECNT and COSECNT FUNCTIONS: These are little less important trigonometric functions. I will just give their epressions and move on. Notation for secant of the angle : sec Definition of secant: sec = cos Notation for cosecant of the angle : csc Definition of cosecant: csc = sin. TRIGONOMETRIC RTIOS OF QUDRNTL NGLES: (DEG) (RD) sin cos tan cot sec csc 0 o 0 0 0 90 o 0 Undefined 80 o 0-0 70 o 3-0 undefined Undefined 0 Undefined 0 Undefined - Undefined Undefined Undefined -
.3 TRIGONOMETRIC FUNCTIONS SSOCITED WITH CUTE NGLES ND SOLVING RIGHT TRINGLE: If is an acute angle, then 0 o,90 o or 0 rad, rad Consider this right triangle. For an acute angle in a right triangle, β c a b C * cos = (length of side adjacent to ) (length of hpotenuse) Hence cos = b c and cos = a c * sin = (length of side opposite to ) (length of hpotenuse) Hence sin = a c and sin =b c * tan = (length of side opposite to ) (length of side adjacent to ) Hence tan = a b and tan = b a * cot = (length of side adjacent to ) (length of side opposite to ) Hence * sec = * csc = cot = b a and cos = a b cos sin Hence sec = c b and sec = c a Hence csc = c a and csc = c b Eample: When the angle of elevation of the sun is 30 o, a tree casts a shadow of 0 m long. How tall is the tree? ns: With the help of the figure, we can write From this we get a=0 tan 30=5.77 m c 60 a=? tan 30= a b. Equivalentl, a=b tan 30 30 b=0 m C
.4 RELTION ETWEEN TRIGONOMETRIC RTIOS OF TWO COMPLEMENTRY NGLES: If two angles are complementar, that is, their sum is equal to 90 o (or radians) then sine, tangent, and secant functions of one angle is equal to the COfunction of the other one. +β=90, therefore, sin = cos β = a / c tan = cot β = a / b sec = csc β = c / b c β a b C In other words, +β=90, therefore, β = 90 - sin β = sin (90 - ) = cos tan β = tan (90 - ) = cot sec β = sec (90 - ) = csc β Eamples: sin 40 o =cos 90 o 40 o =cos50 o sec89=csc 90 o 89 o =csc o cot 3 8 =tan 3 8 =tan 8 tan 0=cot 90 o 0 o =cot 80 o cos =sin 90 o o =sin 88 o cos 5 =sin 5 =cos We can obtain the values of trigonometric functions for 30 o, 45 o and 60 o b using trigonometric ratios in a right triangle.
.5 TRIGONOMETRIC RTIOS OF 30 o, 45 o,60 o The values of these functions for 30 o and 60 o can be evaluated b drawing an equilateral triangle with side length, and for 45 o b drawing a square with side length. First, let us use the equilateral triangle of side length below to determine the trigonometric ratios of 30 o and 60 o C Let us draw the altitude to one of the sides to bisect it: C 30 60 / D / Namel we obtain a 30 o 60 o 90 o right triangle DC. It remains to determine its sides. The hpotenuse and one of the right sides is known. Therefore we can use the pthagorean theorem to determine the unknown side: Using C = D CD, we can write = CD CD = 4 = 3 4. Namel CD = 3.
C 30 3/ 60 / D / sin 30= / =/ sin 60= 3/ = 3/ cos30= 3/ = 3/ cos60= / =/ tan 30= / 3 =/ 3 tan 60= 3/ = 3 cot 30= 3/ / = 3 cot 60= 3 =/ 3 Remark: For an two complementar angles, the values of the functions sine and tangent of one angle is equal to the co-function values of the other angle. Namel: sin 30=cos60,sin 60=cos 30, tan 30=cot 60, tan 60=cot 30. Net, let us use the square of side length below to determine the trigonometric ratios of 45 o D C
Let us draw one of the diagonals to divide the square into two right triangles: D C 45 45 Namel we obtain a 45 o 45 o 90 o right triangle C. It remains to determine its sides. The right sides are known. Therefore we can use the pthagorean theorem to determine the hpotenuse: Using C = C, we can write C = = C =. D C 45 45 sin 45= cos 45= tan 45= = cot 45= = Let's repeat the remark about complementar angles: For an two complementar angles, the values of the functions sine and tangent of one angle is equal to the cofunction values of the other angle. Namel sin 45=cos 45, tan 45=cot 45. Now, let us summarize these results together in a single table:
C D C 30 3/ 45 60 / D / 45 sin 30= / =/ sin 60= 3/ = 3/ sin 45= cos 30= 3/ = 3/ cos60= / =/ cos 45= tan 30= / 3 =/ 3 tan 60= 3/ = 3 tan 45= = cot 30= 3/ = 3 cot 60= / 3 =/ 3 cot 45= =.6 FUNDMENTL TRIGONOMETRIC IDENTITIES: IDENTITY ONE: For all R, sin cos = We proved this identit using a unit circle. Now let us prove it using a right triangle: sin = a / c sin = a / c cos = b / c cos = b / c sin + cos = (a / c ) + (b / c ) sin + cos = (a + b ) / c pthagorean identit, a + b = c sin + cos = c / c = b c β a C
Conclusion: sin = cos sin =± cos, cos = sin cos =± sin Eample: Calculate sin cos if sin cos = 3 nswer: Squaring the difference, we get an equation sin cos sin cos =. 9 Using the identit one, the first two terms simplif to. Hence the equation becomes sin cos = 9 IDENTITY TWO: from which we get sin cos = 9 = 8 9 sin cos = 4 9 tan = sin cos cos, cot = sin It it ver eas to derive this using a right triangle. Let's do it with unit circle. Let's work with the figure below: Just use the definition of tangent and cotangent. ais (or sine ais) Q (-,0) C(0,) cot P cos sin M R tan (,0) cotangent ais ais (or cosine ais) P ( cos, sin ) Q (, tan ) R ( cot, ) (0,-) tangent ais We will use similarit of triangles MP and Q first: Namel tan = sin cos Q = MP M tan = sin cos. Net we use similarit of triangles MP and CR: CR C = M MP cot = cos sin. Namel cot = cos sin. Conclusion: tan cot =
Eample: Calculate tan cot if tan cot =4 nswer: Here we can use a shortcut: = 4. Namel tan cot = tan cot 4 tan cot Using the given information and the identit two, we get tan cot =4 4 =6 4=0 Taking the square root, we get tan cot =± 0 IDENTITY THREE: sec = cos, csc = sin Eample: Prove that tan =sec Here use the identit two to write tan = sin cos = cos cos sin cos = cos sin cos The numerator equals (b identit one) hence the equalit simplifies to cos =sec.7 TRIGONOMETRIC VLUES OF Let %alpha be an acute angle. Since the angle is in Quadrant IV, we can write the following rules (just b using the definition of sin, cos, tan, and cot) sin (- ) = - sin ( ) P (sin( ), cos( )) cos (- ) = cos ( ) - tan (- ) = - tan ( ) - cot (- ) = - cot ( ) - P' (sin(- ), cos(- )) Remark: Sine, tangent, and cotangent are odd functions whereas cosine is an even function.
Eample: sin (-60) + cos (-45) + tan (-30) + cot (-45) = sin 60 cos 45 tan 30 cot 45 = 3 3.8 FINDING THE REMINING TRIGONOMETRIC RTIOS PROVIDED THT ONE OF THEM IS KNOWN Eample: Let 0, and sin = 3 5. Find cos and tan. Solution:. Draw a suitable right triangle such tha sin = 3 / 5. Use pthagorean theore m to de te rmine the unknown side: c = a + b b = c - a = 5-9 = 6 b = 4 3. Use de finition of cosine, tange nt to find cos and tan : cos = b / c = 4 / 5 and tan = a / b = 3 / 4 c=5 b=unknown a=3 C Eample: Let, and cos = 5 3. Find sin, tan and cot. Remark: is not an acute angle. Le t's assume it is an acute angle and we will change the answe r late r accordingl.. Le t us assume is an acute angle whence cos = 5 / 3. Draw a suitable right triangle such that cos = 5 / 3 3. Use pthagorean theore m to de te rmine the unknown side: c = a + b a = c - b = 69-5 = 44 a = 4. Use de finition of sine, tange nt, cotange nt: sin = a / c = / 3, tan = a / b = / 5 and cot = b / a = 5 / 5. In Quadrant II, sine is positiv e while cosine, tange nt and cotange nt are negative (se e section.) There fore, the final answe r is: sin = / 3, tan = - / 5 and cot = - 5 / c=3 b=5 a=? C
.9 SUMMRY Let's summarize section two with a big unit circle that includes all the multiples of 30 o and 45 o angles. nalze the figure below carefull: (- C (- 3, ) D (-, 3 ), ) 35 50 0 (0,) 0.8 0.6 0.4 90 60 D (, 3 ) 45 C ( 30, ) ( 3, ) 0. 3 (-,0) 80 0 (,0) 360 -.5 - -0.5 0.5.5-0. 3 (- C 3 (- 3,- ) 0,- ) 5 D 3 (-,- 3 ) 40-0.4-0.6-0.8 70-4 (0,-) 300 35 330 C 4 ( D 4 (,- 3 ) 4 ( 3,- ),- )