Trigonometric Functions
Unit Circle
The unit circle is the circle U with center $ (0,0) $ and radius 1. The equation of the unit circle is $ x^{2} + y^{2} = 1 $ . The circumference of the unit circle is $ 2$ .
EXAMPLE 20.1 Draw a unit circle and indicate its intercepts (see Fig. 20-1).
Points on a Unit Circle
A unique point P on a unit circle U can be associated with any given real number t in the following manner:
Associated with t = 0 is the point (1,0).
Associated with any positive real number t is the point $ P(x,y) $ found by proceeding a distance $ |t| $ in the counterclockwise direction from the point $ (1,0) $ (see Fig. 20-2).
Associated with any negative real number t is the point $ P(x,y) $ found by proceeding a distance $ |t| $ in the clockwise direction from the point $ (1,0) $ (see Fig. 20-3).
Definition of the Trigonometric Functions
If t is a real number and $ P(x,y) $ is the point, referred to as $ P(t) $ , on the unit circle U that corresponds to P, then the six trigonometric functions of t—sine, cosine, tangent, coscat, secant, and cotangent, abbreviated sin, cos, tan, csc, sec, and cot, respectively—are defined as follows:
\[ \sin t=y \]
\[ \csc t=\frac{1}{y}\left(if y\neq0\right) \]
\[ \cos t=x \]
\[ \sec t=\frac{1}{x}\left(if x\neq0\right) \]
\[ \tan t=\frac{y}{x}\left(if x\neq0\right) \]
\[ \cot t=\frac{x}{y}\left(if y\neq0\right) \]
EXAMPLE 20.2 If t is a real number such that $ P(, -) $ is the point on the unit circle that corresponds to t, find the six trigonometric functions of t.
Since the x coordinate of P is $ $ and the y coordinate of P is $ - $ , the six trigonometric functions of t are as follows:
\[ \sin t=y=-\frac{4}{5} \]
\[ \cos t=x=\frac{3}{5} \]
\[ \tan t=\frac{y}{x}=\frac{-4/5}{3/5}=-\frac{4}{3} \]
\[ \csc t=\frac{1}{y}=\frac{1}{-4/5}=-\frac{5}{4} \]
\[ \sec t=\frac{1}{x}=\frac{1}{3/5}=\frac{5}{3} \]
\[ \cot t=\frac{x}{y}=\frac{3/5}{-4/5}=-\frac{3}{4} \]
Symmetries of the Points on a Unit Circle
For any real number t, the following relations can be shown to hold:
$ P(t + 2) = P(t) $ .
If $ P(t) = (x,y) $ , then $ P(-t) = (x,-y) $ .
If $ P(t) = (x, y) $ , then $ P(t + ) = (-x, -y) $ .
Periodic Functions
A function f is called periodic if there exists a real number p such that $ f(t + p) = f(t) $ for every real number t in the domain of f. The smallest such real number is called the period of the function.
Periodicity of the Trigonometric Functions
The trigonometric functions are all periodic. The following important relations can be shown to hold:
\[ \sin(t+2\pi)=\sin t \]
\[ \cos(t+2\pi)=\cos t \]
\[ \csc(t+2\pi)=\csc t \]
\[ \tan(t+\pi)=\tan t \]
\[ \sec(t+2\pi)=\sec t \]
\[ \cot(t+\pi)=\cot t \]
Notation
Notation for exponents: The expressions for the squares of the trigonometric functions arise frequently. $ (t)^{2} $ is generally written $ ^{2}t $ , $ (t)^{2} $ is generally written $ ^{2}t $ , and so on. Similarly, $ (t)^{3} $ is generally written $ ^{3}t $ , and so on.
I dentities
An identity is an equation that is true for all values of the variables it contains, as long as both sides are meaningful.
Trigonometric Identities
- PYTHAGOREAN IDENTITIES. For all t for which both sides are defined:
\[ \cos^{2}t+\sin^{2}t=1 \]
\[ 1+\tan^{2}t=\sec^{2}t \]
\[ \cot^{2}t+1=\csc^{2}t \]
\[ \cos^{2}t=1-\sin^{2}t \]
\[ \tan^{2}t=\sec^{2}t-1 \]
\[ \cot^{2}t=\csc^{2}t-1 \]
\[ \sin^{2}t=1-\cos^{2}t \]
\[ 1=\sec^{2}t-\tan^{2}t \]
\[ 1=\csc^{2}t-\cot^{2}t \]
- RECIPROCAL IDENTITIES. For all t for which both sides are defined:
\[ \sin t=\frac{1}{\csc t} \]
\[ \cos t=\frac{1}{\sec t} \]
\[ \tan t=\frac{1}{\cot t} \]
\[ \csc t=\frac{1}{\sin t} \]
\[ \sec t=\frac{1}{\cos t} \]
\[ \cot t=\frac{1}{\tan t} \]
- QUOTIENT IDENTITIES. For all t for which both sides are defined:
\[ \tan t=\frac{\sin t}{\cos t} \]
\[ \cot t=\frac{\cos t}{\sin t} \]
- IDENTITIES FOR NEGATIVES. For all t for which both sides are defined:
\[ \sin(-t)=-\sin t \]
\[ \cos(-t)=\cos t \]
\[ \csc(-t)=-\csc t \]
\[ \sec(-t)=\sec t \]
\[ \tan(-t)=-\tan t \]
\[ \cot(-t)=-\cot t \]
SOLVED PROBLEMS
20.1. Find the domain and range of the sine and cosine functions.
For any real number t, a unique point $ P(t) = (x, y) $ on the unit circle $ x^{2} + y^{2} = 1 $ is associated with t. Since $ t = y $ and $ t = x $ are defined for all t, the domain of the sine and cosine functions is R. Since y and x are coordinates of points on the unit circle, $ -1 y $ and $ -1 x $ , hence the range of the sine and cosine functions is given by $ -1 t $ and $ -1 t $ , that is, [-1, 1].
20.2. For what values of t is the y-coordinate of $ P(t) $ equal to zero?
See Fig. 20-5.
By definition, $ P(0) = (1,0) $ . Since the perimeter of the unit circle is $ 2$ , if t is any positive or negative integer multiple of $ 2$ , then again $ P(t) = (1,0) $ .
Since $ $ is half the perimeter of the unit circle, $ P() $ is halfway around the unit circle from $ (1,0) $ ; that is, $ P() = (-1,0) $ . Furthermore, if t is equal to $ $ plus any positive or negative integer multiple of $ 2$ , then again $ P(t) = (-1,0) $ .
Summarizing, the y-coordinate of $ P(t) $ is equal to zero if t is any integer multiple of $ $ ; thus, $ n$ .
20.3. For what values of t is the x-coordinate of $ P(t) $ equal to zero?
See Fig. 20-6. Since the perimeter of the unit circle is $ 2$ , one-fourth of the perimeter is $ /2 $ . Thus $ P(/2) $ is one-fourth of the way around the unit circle from $ (1,0) $ ; that is, $ P(/2) = (0,1) $ . Also, if t is equal to $ /2 $ plus any positive or negative integer multiple of $ 2$ , then again $ P(t) = (0,1) $ .
Next, if $ t = + /2 $ , or $ 3/2 $ , then t is three-fourths of the way around the unit circle from $ (1,0) $ ; that is, $ P(3/2) = (0, -1) $ . And if t is equal to $ 3/2 $ plus any positive or negative integer multiple of $ 2$ , then again $ P(t) = (0, -1) $ .
Summarizing, the x-coordinate of $ P(t) $ is equal to zero if t is $ /2 $ or $ 3/2 $ plus any integer multiple of $ 2$ ; thus, $ /2 + 2n $ or $ 3/2 + 2n $ .
20.4. Find the domains of the tangent and secant functions
For any real number t, a unique point $ P(t) = (x, y) $ on the unit circle $ x^{2} + y^{2} = 1 $ is associated with t. Since $ t $ is defined as y/x and sect is defined as 1/x, each function is defined for all values of t except those for which x = 0. From Problem 20.3, these values are $ /2 + 2n $ or $ 3/2 + 2n $ , n any integer. Thus, the domains of the tangent and secant functions are $ {t R | t /2 + 2n, 3/2 + 2n} $ , n any integer.
20.5. Find the domains of the cotangent and cosecant functions.
For any real number t, a unique point $ P(t) = (x, y) $ on the unit circle $ x^{2} + y^{2} = 1 $ is associated with t. Since $ t $ is defined as x/y and $ t $ is defined as 1/y, each function is defined for all values of t except those for which y = 0. From Problem 20.2, these values are $ n$ , for n any integer. Thus, the domains of the cotangent and cosecant functions are $ {t R | t n} $ , n any integer.
20.6 Find the ranges of the tangent, cotangent, secant, and cosecant functions
For any real number t, a unique point $ P(t) = (x, y) $ on the unit circle $ x^{2} + y^{2} = 1 $ is associated with t. Since $ t $ is defined as y/x and $ t $ is defined as x/y, and, since, for various values of t, x may be greater than y, less than y, or equal to y, $ t = y/x $ and $ t = x/y $ may assume any real value. Thus, the ranges of the tangent and cotangent functions are both R.
Since $ t $ is defined as 1/x and $ t $ is defined as 1/y, and for any point on the unit circle, $ -1 x $ and $ -1 y $ , it follows that $ |l/x| $ and $ |l/y| $ , that is, $ |t| $ and $ |t| $ . Thus, the ranges of the secant and cosecant functions are both $ (-, -1] (extract/viewrange_chunk_1_1_5_53f7d0aa/images/page_5_img_in_image_box_439_378_702_599.jpg)
Since $ P(0) = (1, 0) = (x, y) $ , it follows that
\[ \sin(0)=y=0 \]
csc(0) = 1/y = 1/0 is undefined
\[ \cos(0)=x=1 \]
\[ \sec(0)=1/x=1/1=1 \]
\[ \tan(0)=y/x=0/1=0 \]
cot(0) = x/y = 1/0 is undefined
20.8. Find the six trigonometric functions of $ /2 $ .
See Fig. 20-8. Since the circumference of the unit circle is $ 2$ , $ P(/2) $ is one-fourth of the way around the unit circle from (1, 0). Thus $ P(/2) = (0, 1) = (x, y) $ and it follows that
\[ \sin(\pi/2)=y=1 \]
\[ \csc(\pi/2)=1/y=1/1=1 \]
\[ \cos(\pi/2)=x=0 \]
\[ \sec(\pi/2) = 1/x = 1/0 is undefined \]
tan(π/2) = y/x = 1/0 is undefined
\[ \cot(\pi/2)=x/y=0/1=0 \]
20.9. If $ P(t) $ is in a quadrant, it is said that t is in that quadrant. For t in each of the four quadrants, derive the following table showing the signs of the six trigonometric functions of t.
| QUADRANT I | QUADRANT II | QUADRANT III | QUADRANT IV | |
| $ t \(</td><td>+</td><td>+</td><td>−</td><td>−</td></tr><tr><td>\) t \(</td><td>+</td><td>−</td><td>−</td><td>+</td></tr><tr><td>\) t \(</td><td>+</td><td>−</td><td>+</td><td>−</td></tr><tr><td>\) t \(</td><td>+</td><td>+</td><td>−</td><td>−</td></tr><tr><td>\) t \(</td><td>+</td><td>−</td><td>−</td><td>+</td></tr><tr><td>\) t $ | + | − | + | − |
Since $ t = y $ and $ t = 1/y $ , and y is positive in quadrants I and II, and negative in quadrants III and IV, the signs of $ t $ and $ t $ are as shown.
Since $ t = x $ and $ t = 1/x $ , and x is positive in quadrants I and IV, and negative in quadrants II and III, the signs of $ t $ and $ t $ are as shown.
Since $ t = y/x $ and $ t = x/y $ , and x and y have the same signs in quadrants I and III and opposite signs in quadrants II and IV, the signs of $ t $ and $ t $ are as shown.
20.10. Find the six trigonometric functions of $ /4 $
See Fig. 20-9. Since $ /4 $ is one-half the way from 0 to $ /2 $ , the point $ P(/4) = (x, y) $ lies on the line y = x. Thus the coordinates $ (x, y) $ satisfy both $ x^{2} + y^{2} = 1 $ and y = x. Substituting yields:
\[ \begin{aligned}x^{2}+x^{2}&=1\\2x^{2}&=1\\x^{2}&=1/2\\x&=1/\sqrt{2}\end{aligned} \]
since x is positive
Hence $ P(/4)=(x,y)=(1/,1/) $ . Hence it follows that:
\[ \sin\left(\frac{\pi}{4}\right)=y=\frac{1}{\sqrt{2}} \]
\[ \csc\left(\frac{\pi}{4}\right)=\frac{1}{y}=\frac{1}{1/\sqrt{2}}=\sqrt{2} \]
\[ \cos\left(\frac{\pi}{4}\right)=x=\frac{1}{\sqrt{2}} \]
\[ \sec\left(\frac{\pi}{4}\right)=\frac{1}{x}=\frac{1}{1/\sqrt{2}}=\sqrt{2} \]
\[ \tan\left(\frac{\pi}{4}\right)=\frac{y}{x}=\frac{1/\sqrt{2}}{1/\sqrt{2}}=1 \]
\[ \cot\left(\frac{\pi}{4}\right)=\frac{x}{y}=\frac{1/\sqrt{2}}{1/\sqrt{2}}=1 \]
20.11. Prove the symmetry properties listed on page 177 for points on a unit circle.
For any real number t, $ P(t + 2) = P(t) $ .
If $ P(t) = (x, y) $ , then $ P(-t) = (x, -y) $ .
If $ P(t) = (x, y) $ , then $ P(t + ) = (-x, -y) $ .
Let $ P(t) = (x, y) $ . Since the circumference of the unit circle is precisely $ 2$ , the point $ P(t + 2) $ is obtained by going exactly once around the unit circle from $ P(t) $ . Thus the coordinates of $ P(t + 2) $ are the same as those of $ P(t) $ .
See Fig. 20-10.
Let $ P(t) = (x, y) $ . Since $ P(t) $ and $ P(-t) $ are obtained by going the same distance around the unit circle from the same point, $ P(0) $ , the coordinates of the two points will be equal in absolute value. The x-coordinates of the two points will be the same; however, since the two points are reflections of each other with respect to the x-axis, the y-coordinates of the points will be opposite in sign. Hence the coordinates of $ P(-t) $ are $ (x, -y) $ .
- See Fig. 20-11.
Let $ P(t) = (x, y) $ . Since $ P(t + ) $ is obtained by going halfway around the unit circle from $ P(t) $ , the two points are at opposite ends of a diameter, hence they are reflections of each other with respect to the origin. Hence $ P(t + ) = (-x, -y) $ .
20.12. Find the six trigonometric functions of 5π/4.
Since $ = + $ , and $ P() = (, ) $ , it follows that $ P() = (-, -) $ . Hence the six trigonometric functions of $ 5/4 $ are:
\[ \sin\left(\frac{5\pi}{4}\right)=y=-\frac{1}{\sqrt{2}}\quad\csc\left(\frac{5\pi}{4}\right)=\frac{1}{y}=\frac{1}{-1/\sqrt{2}}=-\sqrt{2} \]
\[ \cos\left(\frac{5\pi}{4}\right)=x=-\frac{1}{\sqrt{2}}\quad\quad\quad\sec\left(\frac{5\pi}{4}\right)=\frac{1}{x}=\frac{1}{-1/\sqrt{2}}=-\sqrt{2} \]
\[ \tan\Big(\frac{5\pi}{4}\Big)=\frac{y}{x}=\frac{-1/\sqrt{2}}{-1/\sqrt{2}}=1\qquad\cot\Big(\frac{5\pi}{4}\Big)=\frac{x}{y}=\frac{-1/\sqrt{2}}{-1/\sqrt{2}}=1 \]
20.13. Prove the periodicity properties for the sine, cosine, and tangent functions.
Let $ P(t) = (x, y) $ ; then $ P(t + 2) = P(t) = (x, y) $ . It follows immediately that $ (t + 2) = y = t $ and $ (t + 2) = x = t $ .
Also, $ P(t + ) = (-x, -y) $ . Hence $ (t + ) = = = t $ .
20.14. Prove the reciprocal identities.
Let $ P(t) = (x, y) $ ; then it follows that:
\[ \csc t=\frac{1}{y}=\frac{1}{\sin t}\qquad\sec t=\frac{1}{x}=\frac{1}{\cos t}\qquad\cot t=\frac{x}{y}=1\div\frac{y}{x}=1\div(\tan t)=\frac{1}{\tan t} \]
Hence it follows by algebra that:
\[ \sin t=\frac{1}{\csc t}\qquad\cos t=\frac{1}{\sec t}\qquad\tan t=\frac{1}{\cot t} \]
20.15. Prove the periodicity properties for the cosecant, secant, and cotangent functions.
Use the reciprocal identities and the periodicity properties for sine, cosine, and tangent.
\[ \csc(t+2\pi)=\frac{1}{\sin(t+2\pi)}=\frac{1}{\sin t}=\csc t \]
\[ \sec(t+2\pi)=\frac{1}{\cos(t+2\pi)}=\frac{1}{\cos t}=\sec t \]
\[ \cot(t+\pi)=\frac{1}{\tan(t+\pi)}=\frac{1}{\tan t}=\cot t \]
20.16. Prove the quotient identities.
Let $ P(t) = (x, y) $ , then it follows that:
\[ \tan t=\frac{y}{x}=\frac{\sin t}{\cos t}\qquad and\qquad\cot t=\frac{x}{y}=\frac{\cos t}{\sin t} \]
20.17. Find the six trigonometric functions of $ 5/2 $ .
Since \(\frac{5\pi}{2}=2\pi+\frac{\pi}{2}\) and \(P\left(\frac{\pi}{2}\right)=(0,1)\), it follows that \(P\left(\frac{5\pi}{2}\right)=(0,1)\) and
\[ \sin(5\pi/2)=y=1 \]
\[ \csc(5\pi/2)=1/y=1/1=1 \]
\[ \cos(5\pi/2)=x=0 \]
$ (5/2)=1/x=1/0 $ is undefined
\[ \tan(5\pi/2)=y/x=1/0is undefined \]
\[ \cot(5\pi/2)=x/y=0/1=0 \]
20.18. Prove the identities for negatives.
Let $ P(t) = (x, y) $ . Then $ P(-t) = (x, -y) $ by the symmetry properties of points on a unit circle. It follows that:
\[ \sin(-t)=-y=-\sin t \]
\[ \csc(-t)=\frac{1}{\sin(-t)}=\frac{1}{-\sin t}=-\frac{1}{\sin t}=-\csc t \]
\[ \cos(-t)=x=\cos t \]
\[ \sec(-t)=\frac{1}{\cos\left(-t\right)}=\frac{1}{\cos t}=\sec t \]
\[ \tan(-t)=\frac{-y}{x}=-\frac{y}{x}=-\tan t \]
\[ \cot(-t)=\frac{x}{-y}=-\frac{x}{y}=-\cot t \]
20.19. Find the six trigonometric functions of $ -/4 $ .
Use the identities for negatives and the results of Problem 20.10.
\[ \sin\left(-\frac{\pi}{4}\right)=-\sin\left(\frac{\pi}{4}\right)=-\frac{1}{\sqrt{2}}\qquad\cos\left(-\frac{\pi}{4}\right)=\cos\left(\frac{\pi}{4}\right)=\frac{1}{\sqrt{2}}\qquad\tan\left(-\frac{\pi}{4}\right)=-\tan\left(\frac{\pi}{4}\right)=-1 \]
\[ \csc\left(-\frac{\pi}{4}\right)=-\csc\left(\frac{\pi}{4}\right)=-\sqrt{2}\qquad\sec\left(-\frac{\pi}{4}\right)=\sec\left(\frac{\pi}{4}\right)=\sqrt{2}\qquad\cot\left(-\frac{\pi}{4}\right)=-\cot\left(\frac{\pi}{4}\right)=-1 \]
20.20. Prove the Pythagorean identity $ ^{2}t + ^{2}t = 1 $ .
For any real number t, a unique point $ P(t) = (x, y) $ on the unit circle $ x^{2} + y^{2} = 1 $ is associated with t. By definition, $ t = x $ and $ t = y $ ; hence for any $ t, (t)^{2} + (t)^{2} = 1 $ , that is,
\[ \cos^{2}t+\sin^{2}t=1 \]
20.21. Prove the Pythagorean identity $ 1 + ^{2}t = ^{2}t $ .
Start with $ ^{2}t + ^{2}t = 1 $ and divide both sides by $ ^{2}t $ . Then it follows that:
\[ \frac{\cos^{2}t}{\cos^{2}t}+\frac{\sin^{2}t}{\cos^{2}t}=\frac{1}{\cos^{2}t} \]
\[ 1+\left(\frac{\sin t}{\cos t}\right)^{2}=\left(\frac{1}{\cos t}\right)^{2} \]
\[ 1+\tan^{2}t=\sec^{2}t \]
20.22. Given $ t = $ and t in quadrant II, find the other five trigonometric functions of t.
- Cosine. From the Pythagorean identity, $ ^{2}t = 1 - ^{2}t $ . Since t is specified in quadrant II, $ t $ must be negative (see Problem 20.9). Hence,
\[ \cos t=-\sqrt{1-\sin^{2}t}=-\sqrt{1-\left(\frac{1}{2}\right)^{2}}=-\sqrt{\frac{3}{4}}=-\frac{\sqrt{3}}{2} \]
- Tangent. From the quotient identity,
\[ \tan t=\frac{\sin t}{\cos t}=\frac{\frac{1}{2}}{-\sqrt{3}/2}=-\frac{1}{\sqrt{3}} \]
- Cotangent. From the reciprocal identity,
\[ \cot t=\frac{1}{\tan t}=\frac{1}{-1/\sqrt{3}}=-\sqrt{3} \]
- Secant. From the reciprocal identity,
\[ \sec t=\frac{1}{\cos t}=\frac{1}{-\sqrt{3}/2}=-\frac{2}{\sqrt{3}} \]
- Cosecant. From the reciprocal identity,
\[ \csc t=\frac{1}{\sin t}=\frac{1}{\frac{1}{2}}=2 \]
20.23. Given $ t = -2 $ and t in quadrant IV, find the other five trigonometric functions of t.
- Secant. From the Pythagorean identity, $ ^{2}t = 1 + ^{2}t $ . Since t is specified in quadrant IV, sec t must be positive (see Problem 20.9). Hence,
\[ \sec t=\sqrt{1+\tan^{2}t}=\sqrt{1+(-2)^{2}}=\sqrt{5} \]
- Cosine. From the reciprocal identity,
\[ \cos t=\frac{1}{\sec t}=\frac{1}{\sqrt{5}} \]
- Sine. From the quotient identity, $ t = $ ; hence,
\[ \sin t=\tan t\cos t=(-2)\frac{1}{\sqrt{5}}=-\frac{2}{\sqrt{5}} \]
- Cotangent. From the reciprocal identity,
\[ \cot t=\frac{1}{\tan t}=\frac{1}{-2}=-\frac{1}{2} \]
- Cosecant. From the reciprocal identity,
\[ \csc t=\frac{1}{\sin t}=\frac{1}{-2/\sqrt{5}}=-\frac{\sqrt{5}}{2} \]
20.24. For an arbitrary value of t express the other trigonometric functions in terms of $ t $ .
Cosine. From the Pythagorean identity, \(\cos^{2}t = 1 - \sin^{2}t\). Hence \(\cos t = \pm \sqrt{1 - \sin^{2}t}\).
Tangent. From the quotient identity, $ t = $ . Using the previous result, $ t = $ .
Cotangent. From the quotient identity, $ t = $ . Hence $ t = $ .
Secant. From the reciprocal identity, $ t = $ . Hence $ t = = $ .
Cosecant. From the reciprocal identity, $ t = $
SUPPLEMENTARY PROBLEMS
20.25. If t is a point on the unit circle with coordinates $ (-, -) $ , find the six trigonometric functions of t.
Ans. $ t = -12/13, t = -5/13, t = 12/5, t = 5/12, t = -13/5, t = -13/12 $
20.26. If t is a point on the unit circle with coordinates $ (, -) $ , find the six trigonometric functions of t.
Ans. $ t = -1/, t = 2/, t = -1/2, t = -2, t = /2, t = - $
20.27. Find the six trigonometric functions of $ $ .
Ans. \(\sin \pi = 0\), \(\cos \pi = -1\), \(\tan \pi = 0\), \(\cot \pi\) is undefined, \(\sec \pi = -1\), \(\csc \pi\) is undefined.
20.28. Find the six trigonometric functions of $ -/2 $ .
Ans. \(\sin(-\pi/2) = -1\), \(\cos(-\pi/2) = 0\), \(\tan(-\pi/2)\) is undefined,
cot(−π/2) = 0, sec(−π/2) is undefined, csc(−π/2) = −1
20.29. Find the six trigonometric functions of 7π/4.
\[ \sin(7\pi/4)=-1/\sqrt{2},\cos(7\pi/4)=1/\sqrt{2},\tan(7\pi/4)=-1, \]
\[ \cot(7\pi/4)=-1,\sec(7\pi/4)=\sqrt{2},\csc(7\pi/4)=-\sqrt{2} \]
20.30. Prove that for all \(t\), \(\sin(t + 2\pi n) = \sin t\) for any integer value of \(n\).
20.31. Prove the Pythagorean identity $ ^{2}t + 1 = ^{2}t $ .
20.32. Given $ t = 2/5 $ and t in quadrant I, find the other five trigonometric functions of t.
\[ Ans.\quad\sin t=\sqrt{21}/5,\tan t=\sqrt{21}/2,\cot t=2/\sqrt{21},\sec t=5/2,\csc t=5/\sqrt{21} \]
20.33. Given $ t = -2/3 $ and t in quadrant IV, find the other five trigonometric functions of t.
\[ Ans.\quad\sin t=-2/\sqrt{13},\cos t=3/\sqrt{13},cot t=-3/2,\sec t=\sqrt{13}/3,csc t=-\sqrt{13}/2 \]
20.34. Given $ t = $ and t in quadrant III, find the other five trigonometric functions of t.
\[ Ans.\quad\sin t=-1/\sqrt{6},\ \cos t=-\sqrt{5}/\sqrt{6},\ \tan t=1/\sqrt{5},\ \sec t=-\sqrt{6}/\sqrt{5},\ \csc t=-\sqrt{6} \]
20.35. Given $ t = - $ and t in quadrant II, find the other five trigonometric functions of t.
\[ Ans.\quad\sin t=\frac{12}{13},\ \cos t=-\frac{5}{13},\ \tan t=-\frac{12}{5},\ \cot t=-\frac{5}{12},\ \csc t=\frac{13}{12} \]
20.36. Given $ t = a $ and t in quadrant II, find the other five trigonometric functions of t.
\[ Ans.\quad\cos t=-\sqrt{1-a^{2}},\tan t=-\frac{a}{\sqrt{1-a^{2}}},\cot t=\frac{-\sqrt{1-a^{2}}}{a},\sec t=-\frac{1}{\sqrt{1-a^{2}}},\csc t=\frac{1}{a} \]
20.37. Given $ t = a $ and t in quadrant IV, find the other five trigonometric functions of t.
\[ Ans.\quad\sin t=-\sqrt{1-a^{2}},\tan t=\frac{-\sqrt{1-a^{2}}}{a},\cot t=-\frac{a}{\sqrt{1-a^{2}}},\sec t=\frac{1}{a},\csc t=-\frac{1}{\sqrt{1-a^{2}}} \]
20.38. Given $ t = a $ and t in quadrant II, find the other five trigonometric functions of t.
\[ Ans.\quad\sin t=-\frac{a}{\sqrt{a^{2}+1}},\cos t=-\frac{1}{\sqrt{a^{2}+1}},\cot t=\frac{1}{a},\sec t=-\sqrt{a^{2}+1},\csc t=\frac{-\sqrt{a^{2}+1}}{a} \]
20.39. For an arbitrary value of t, express the other trigonometric functions in terms of tan t.
\[ Ans.\quad\sin t=\pm\frac{\tan t}{\sqrt{1+\tan^{2}t}},\quad\cos t=\pm\frac{1}{\sqrt{1+\tan^{2}t}},\quad\cot t=\frac{1}{\tan t}, \]
\[ \sec t=\pm\sqrt{1+\tan^{2}t},\csc t=\pm\frac{\sqrt{1+\tan^{2}t}}{\tan t} \]
20.40. For an arbitrary value of t, express the other trigonometric functions in terms of cost.
\[ Ans.\quad\sin t=\pm\sqrt{1-\cos^{2}t},\quad\tan t=\pm\frac{\sqrt{1-\cos^{2}t}}{\cos t},\quad\cot t=\pm\frac{\cos t}{\sqrt{1-\cos^{2}t}}, \]
\[ \sec t=\frac{1}{\cos t},\csc t=\pm\frac{1}{\sqrt{1-\cos^{2}t}} \]
20.41. Show that cosine and secant are even functions.
20.42. Show that sine, tangent, cotangent, and cosecant are odd functions.