Trigonometric Identities and Equations

Definition of Identity

An identity is a statement that two quantities are equal that is true for all values of the variables for which the statement is meaningful.

EXAMPLE 23.1 Which of the following statements is an identity?

\[ \left(a\right)x+3=3+x;\left(b\right)x+3=5;\left(c\right)x\cdot\frac{1}{x}=1. \]

  1. is an identity since it is always true; (b) is not an identity since it is true only if x = 2; (c) is an identity since it is true unless x = 0, in which case it is not meaningful.

Basic Trigonometric Identities

Basic trigonometric identities are repeated below for reference:

  1. Pythagorean Identities. For all t for which both sides are defined:

\[ \cos^{2}t+\sin^{2}t=1 \]

\[ 1+\tan^{2}t=\sec^{2}t\qquad\cot^{2}t+1=\csc^{2}t \]

\[ \cos^{2}t=1-\sin^{2}t \]

\[ \tan^{2}t=\sec^{2}t-1\qquad\cot^{2}t=\csc^{2}t-1 \]

\[ \sin^{2}t=1-\cos^{2}t\qquad1=\sec^{2}t-\tan^{2}t\qquad1=\csc^{2}t-\cot^{2}t \]

  1. RECIPROCAL IDENTITIES. For all t for which both sides are defined:

\[ \sin t=\frac{1}{\csc t}\qquad\cos t=\frac{1}{\sec t}\qquad\tan t=\frac{1}{\cot t} \]

\[ \csc t=\frac{1}{\sin t}\qquad\sec t=\frac{1}{\cos t}\qquad\cot t=\frac{1}{\tan t} \]

  1. QUOTIENT IDENTITIES. For all t for which both sides are defined:

\[ \tan t=\frac{\sin t}{\cos t}\qquad\cot t=\frac{\cos t}{\sin t} \]

  1. IDENTITIES FOR NEGATIVES. For all t for which both sides are defined:

\[ \sin(-t)=-\sin t\qquad\cos(-t)=\cos t\qquad\tan(-t)=-\tan t \]

\[ \csc(-t)=-\csc t\qquad\sec(-t)=\sec t\qquad\cot(-t)=-\cot t \]

Simplifying Trigonometric Expressions

The basic trigonometric identities are used to reduce trigonometric expressions to simpler form:

EXAMPLE 23.2 Simplify: $ $ From the Pythagorean identity, $ 1 - ^{2}= ^{2}$ . Hence, $ = = $ .

V erifying Trigonometric Identities

To verify that a given statement is an identity, show that one side can be transformed into the other by using algebraic techniques, including simplification and substitution, and trigonometric techniques, frequently including reducing other functions to sines and cosines.

EXAMPLE 23.3 Verify that $ (1 - )(1 + ) = ^{2}$ is an identity.

Starting with the left side, an obvious first step is to perform algebraic operations:

\[ \begin{aligned}(1-\cos\theta)(1+\cos\theta)&=1-\cos^{2}\theta\quad Algebra\\&=\sin^{2}\theta\quad Pythagorean identity\end{aligned} \]

EXAMPLE 23.4 Verify that $ = ^{2} t $ is an identity.

Starting with the left side, an obvious first step is to reduce to sines and cosines:

\[ \begin{aligned}\frac{\sin t\cos t}{\tan t}&=\frac{\sin t\cos t}{\sin t/\cos t}&Quotient identity\\ &=\sin t\cos t\div\frac{\sin t}{\cos t}&Algebra\\ &=\sin t\cos t\cdot\frac{\cos t}{\sin t}&Algebra\\ &=\cos^{2}t&Algebra\\ \end{aligned} \]

Nonidentity Statements

If a statement is meaningful yet not true for even one value of the variable or variables, it is not an identity. To show that it is not an identity, it is sufficient to find one value of the variable or variables that would make it false.

EXAMPLE 23.5 Show that $ t + t = 1 $ is not an identity.

Although this statement is true for some values of t, for example t = 0, it is not an identity. For example, choose $ t = /4 $ . Then

\[ \sin\frac{\pi}{4}+\cos\frac{\pi}{4}=\frac{1}{\sqrt{2}}+\frac{1}{\sqrt{2}}=\frac{2}{\sqrt{2}}=\sqrt{2}\ne1 \]

I nverses of Trigonometric Functions

The trigonometric functions are periodic, and therefore are not one-to-one. However, in the first quadrant, sine and tangent are increasing functions and cosine is decreasing; hence, in this region the functions are one-to-one and thus have inverses. For present purposes, the following notation is used:

\[ t=\sin^{-1}a\ (read:inverse sine of a)\ if\ 0\leq t\leq\pi/2\ and\ \sin t=a. \]

\[ t=\cos^{-1}a\ (read:inverse cosine of a)\ if\ 0\leq t\leq\pi/2\ and\ cost=a. \]

$ t = ^{-1} a $ (read: inverse tangent of a) if $ 0 t < /2 $ and $ t = a $ .

A complete treatment of inverse trigonometric functions is given in Chapter 25.

EXAMPLE 23.6 Find (a) $ ^{-1} $ ; (b) $ ^{-1}0 $ (c) $ ^{-1}1 $ .

  1. There is exactly one value of t such that $ t = /2 $ and $ 0 t /2 $ , that is, $ /3 $ . Hence, $ ^{-1} = $ .

  2. There is exactly one value of t such that $ t = 0 $ and $ 0 t /2 $ , that is, $ /2 $ . Hence, $ ^{-1}0 = $ .

  3. There is exactly one value of t such that $ t = 1 $ and $ 0 t < /2 $ , that is, $ /4 $ . Hence, $ ^{-1}1 = $ .

Trigonometric Equations

Trigonometric equations can be solved by a mixture of algebraic and trigonometric techniques, including reducing other functions to sines and cosines, substitution from known trigonometric identities, algebraic simplification, and so on.

  1. BASIC TRIGONOMETRIC EQUATIONS are equations of the form $ t = a $ , $ t = b $ , $ t = c $ . These are solved by using inverses of trigonometric functions to express all solutions in the interval $ [0, 2) $ and then extending to the entire set of solutions. Some problems, however, specify that only solutions in the interval $ [0, 2) $ are to be found.

  2. OTHER TRIGONOMETRIC EQUATIONS are solved by reducing to basic equations using algebraic and trigonometric techniques.

EXAMPLE 23.7 Find all solutions of $ t = $

First find all solutions in the interval $ [0, 2) $ : Start with

\[ t=\cos^{-1}\frac{1}{2}=\frac{\pi}{3} \]

Since cosine is positive in quadrant I and IV, there is also a solution in quadrant IV with reference angle $ /3 $ , namely $ 2- /3 = 5/3 $ .

Extending to the entire real line, since cosine is periodic with period $ 2$ , all solutions can be written as $ /3 + 2n $ , $ 5/3 + 2n $ , n any integer.

EXAMPLE 23.8 Find all solutions in the interval $ [0, 2) $ for $ 5 t = 3 t - 2 $ .

First reduce this to a basic trigonometric equation by isolating the quantity $ t $ .

\[ \begin{aligned}2\tan t&=-2\\ \tan t&=-1\end{aligned} \]

Now find all solutions of this equation in the interval $ [0, 2) $ . Start with $ ^{-1}1 = /4 $ . Since tangent is negative in quadrants II and IV, the solutions are the angles in these quadrants with reference angle $ /4 $ . These are $ - /4 = 3/4 $ and $ 2- /4 = 7/4 $ .

SOLVED PROBLEMS

23.1. Verify that $ t - t = t t $ is an identity.

Starting with the left side, an obvious first step is to reduce to sines and cosines:

\[ \csc t-\sin t=\frac{1}{\sin t}-\sin t\qquad Reciprocal identity \]

\[ \begin{aligned}c s c t-\sin t&=\frac{1-\sin^{2}t}{\sin t}\quad&Algebra\\ &=\frac{\cos^{2}t}{\sin t}\quad&Pythagorean identity\\ &=\frac{\cos t}{\sin t}\cdot\cos t\quad&Algebra\\ &=c o t t\cos t\quad&Quotient identity\end{aligned} \]

23.2. Verify that $ ^{4}- ^{4}= ^{2}- ^{2}$ is an identity.

Starting with the left side, an obvious first step is to express the fourth powers in terms of squares:

\[ \begin{aligned}\sin^{4}\theta-\cos^{4}\theta&=(\sin^{2}\theta)^{2}-(\cos^{2}\theta)^{2}\quad&Algebra\\&=(\sin^{2}\theta-\cos^{2}\theta)(\sin^{2}\theta+\cos^{2}\theta)\quad&Algebra\\&=(\sin^{2}\theta-\cos^{2}\theta)(1)\quad&Pythagorean identity\\&=\sin^{2}\theta-\cos^{2}\theta\quad&Algebra\end{aligned} \]

23.3. Verify that $ + = 2^{2}x $ is an identity.

Starting with the left side, an obvious first step is to combine the two fractional expressions into one:

\[ \begin{aligned}\frac{1}{1-\cos x}+\frac{1}{1+\cos x}&=\frac{(1+\cos x)+(1-\cos x)}{(1-\cos x)(1+\cos x)}\\&=\frac{2}{1-\cos^{2}x}\end{aligned} \]

Now apply a Pythagorean identity:

\[ \begin{aligned}\frac{2}{1-\cos^{2}x}&=\frac{2}{\sin^{2}x}\quad&Pythagorean identity\\&=2\csc^{2}x\quad&Reciprocal identity\end{aligned} \]

23.4. Verify that $ = $ is an identity.

Often in this context squares of sines and cosines are easier to work with than the functions themselves. Starting with the right side, it is effective to multiply numerator and denominator by the expression $ 1 - $ . (This is analogous to the operations in rationalizing the denominator.)

\[ \begin{aligned}\frac{\sin\theta}{1+\cos\theta}&=\frac{\sin\theta(1-\cos\theta)}{(1+\cos\theta)(1-\cos\theta)}\quad&Algebra\\ &=\frac{\sin\theta(1-\cos\theta)}{1-\cos^{2}\theta}\quad&Algebra\\ &=\frac{\sin\theta(1-\cos\theta)}{\sin^{2}\theta}\quad&Pythagorean ide\\ &=\frac{1-\cos\theta}{\sin\theta}\quad&Algebra\end{aligned} \]

23.5. Verify that $ = $ is an identity.

Starting with the left side, reduce to sines and cosines, then simplify the complex fraction that results by multiplying numerator and denominator by $ x y $ , the LCD of the internal fractions:

\[ \begin{aligned}\frac{\tan x+\tan y}{1-\tan x\tan y}&=\frac{\frac{\sin x}{\cos x}+\frac{\sin y}{\cos y}}{1-\frac{\sin x\sin y}{\cos x\cos y}}&Quotient identity\\ &=\frac{\frac{\sin x}{\cos x}+\frac{\sin y}{\cos y}}{1-\frac{\sin x\sin y}{\cos x\cos y}}\cdot\frac{\cos x\cos y}{\cos x\cos y}&Algebra\\ &=\frac{\sin x\cos y+\cos x\sin y}{\cos x\cos y-\sin x\sin y}&Algebra\end{aligned} \]

23.6. Show that $ = t $ is not an identity.

From the Pythagorean identity, the statement is generally true if the left and right sides have the same sign. To show that it is not an identity, choose a value of t for which $ t $ is negative, for example, $ t = 3/2 $ . Then $ = = 1 $ but $ /2 = -1 $ .

23.7. Show that $ (+ )^{2} = ^{2}+ ^{2}$ is not an identity.

This statement arises from the very common algebraic error of confusing $ (a + b)^{2} $ with $ a^{2} + b^{2} $ . To show that it is not an identity, choose any value of $ $ for which neither $ $ nor $ $ is zero, for example, $ = /6 $ .

Then $ (+){2}=(+){2}= $ but $ {2}+{2}=1 $ (from the Pythagorean identity).

23.8. Simplify the expression $ $ by making the substitution $ x = 5 u $ , $ - u $ .

Making the substitution and factoring the expression under the radical yields an expression that can be simplified by applying a Pythagorean identity:

\[ \sqrt{25-x^{2}}=\sqrt{25-(5\sin u)^{2}}=\sqrt{25-25\sin^{2}u}=\sqrt{25(1-\sin^{2}u)}=\sqrt{25\cos^{2}u}=5\left|\cos u\right| \]

The last expression can be further simplified by observing that the restriction $ - u $ confines u to quadrants I and IV in which $ $ is never negative. In this region, $ 5 | u | = 5 u $ .

23.9. Simplify the expression $ $ by making the substitution $ x = 4 u $ , $ - < u < $ .

Proceed as in the previous problem:

\[ \frac{1}{\sqrt{16+x^{2}}}=\frac{1}{\sqrt{16+(4\tan u)^{2}}}=\frac{1}{\sqrt{16+16\tan^{2}u}}=\frac{1}{\sqrt{16(1+\tan^{2}u)}}=\frac{1}{\sqrt{16\sec^{2}u}}=\frac{1}{4\left|\sec u\right|} \]

The last expression can be further simplified by observing that the restriction $ - < u < $ confines u to quadrants I and IV, in which sec u is never negative. In this region, $ = = $ .

23.10. Find all solutions for $ t = $ .

For this basic trigonometric equation, begin by finding all solutions in the interval \([0, 2\pi)\). Start with

\[ t=\sin^{-1}\frac{\sqrt{3}}{2}=\frac{\pi}{3} \]

Since sine is positive in quadrants I and II, there is also a solution in quadrant II with reference angle $ /3 $ , namely, $ - /3 = 2/3 $ .

Extending to the entire real line, since sine is periodic with period $ 2$ , all solutions can be written as $ /3 + 2n $ , $ 2/3 + 2n $ , n any integer.

23.11. Find all solutions for $ 3 - 4 ^{2} = 0 $

First reduce this to a basic trigonometric equation by isolating the quantity $ $ .

\[ \cos^{2}\theta=\frac{3}{4} \]

\[ \cos\theta=\pm\frac{\sqrt{3}}{2} \]

Since $ ^{-1}= $ , there are four solutions in the interval $ [0,2) $ , namely, $ $ (positive cosine), $ -= $ (negative cosine), $ += $ (negative cosine), and $ 2-= $ (positive cosine). Extending to the entire real line, since cosine is periodic with period $ 2$ , all solutions can be written as $ /6+2n $ , $ 5/6+2n $ , $ 7/6+2n $ , $ 11/6+2n $ , n any integer.

23.12. Find all solutions for $ 22x - 1 = 0 $

Reducing to a basic trigonometric equation yields $ 2x= $ . To solve this, begin with $ 2x=^{-1} $ . Thus, in the interval $ [0,2) $ , $ 2x= $ and $ 2x=2-= $ ; extending to the entire real line yields $ 2x=+2n $ and $ 2x=+2n $ . Hence, isolating x, all solutions are given by $ x=+n $ , $ +n $ , n any integer.

23.13. Find all solutions on the interval $ [0, 2) $ for $ 2 ^{2}u + u = 0 $

This is an equation in quadratic form in the quantity $ u $ . It is most efficiently solved by factoring (alternatively: make the substitution $ v = u $ ):

\[ \begin{aligned}\sin u(2\sin u+1)&=0\\\sin u=0\quad or\quad2\sin u+1&=0\\\sin u&=-\frac{1}{2}\end{aligned} \]

$ u = 0 $ has solutions 0 and $ $ on the interval $ [0, 2) $ . $ u = - $ has solutions $ $ and $ $ on the interval.

Solutions: $ 0, , , $

23.14. Find all angles in the interval $ [0^{}, 360^{}) $ that satisfy $ 2 ^{2}= 1 - $

First, use a Pythagorean identity to reduce to one trigonometric function:

\[ 2(1-\cos^{2}\theta)=1-\cos\theta \]

This is an equation in quadratic form in \(\cos\theta\). Reduce to standard form:

\[ 2-2\cos^{2}\theta=1-\cos\theta \]

\[ 2\cos^{2}\theta-\cos\theta-1=0 \]

This is most efficiently solved by factoring:

\[ (2\cos\theta+1)(\cos\theta-1)=0 \]

\[ \begin{aligned}2\cos\theta+1&=0\quad&or\quad&\cos\theta-1&=0\\\cos\theta&=-\frac{1}{2}\quad&\cos\theta&=1\end{aligned} \]

$ = - $ has solutions in quadrants II and III. Since $ ^{-1} = = 60^{} $ , in the interval $ [0^{}, 360^{}) $ the required angles are $ 180^{} - 60^{} = 120^{} $ and $ 180^{} + 60^{} = 240^{} $ . The only solution of $ = 1 $ in the interval is $ 0^{} $ .

Solutions: $ 0^{} $ , $ 120^{} $ , $ 240^{} $

23.15. Find all solutions on the interval $ [0, 2) $ for $ x + x = 1 $

As remarked in Problem 23.4, squares of sines and cosines are often easier to work with than the functions themselves. Isolate $ x $ and square both sides. Recall (Chapter 5) that raising both sides of an equation to an even power is permissible if all solutions to the resulting equation are checked to see whether they are solutions of the original equation.

\[ \sin x=1-\cos x \]

\[ \begin{aligned}\sin^{2}x&=(1-\cos x)^{2}\\&=1-2\cos x+\cos^{2}x\end{aligned} \]

Now use a Pythagorean identity to reduce to one trigonometric function:

\[ 1-\cos^{2}x=1-2\cos x+\cos^{2}x \]

This is an equation in quadratic form in $ x $ . Reduce to standard form:

\[ 0=2\cos^{2}x-2\cos x \]

Solve by factoring:

\[ \begin{aligned}&2\cos x\left(\cos x-1\right)=0\\&2\cos x=0\quad or\quad\cos x-1=0\\&\cos x=0\quad\cos x=1\end{aligned} \]

On the interval $ [0, 2) $ , $ x = 0 $ has solutions $ $ and $ $ ; $ x = 1 $ has solution 0. It is necessary to check each of these solutions in the original equation:

\[ \begin{aligned}Check:x=0:\sin0+\cos0=1?&\quad x=\frac{\pi}{2}:\sin\frac{\pi}{2}+\cos\frac{\pi}{2}=1?&\quad x=\frac{3\pi}{2}:\sin\frac{3\pi}{2}+\cos\frac{3\pi}{2}=1?\\ &0+1=1\quad1+0=0\quad-1+0\neq1\end{aligned} \]

A solution

A solution

Not a solution

Solutions: $ 0, $

23.16. Find approximate values for all solutions to $ ^{2}t - t - 6 = 0 $ .

This is an equation in quadratic form in the quantity tan t. It is most efficiently solved by factoring:

\[ \begin{aligned}\tan^{2}t-\tan t-6&=0\ $ \tan t-3)(\tan t+2)&=0\\\tan t-3&=0\quad\quad or\quad\quad\tan t+2=0\\\tan t&=3\quad\quad\quad\tan t=-2\end{aligned} \]

Use the calculator to find approximate values for solutions of these equations: $ t = 3 $ has solutions in quadrants I and III; since $ ^{-1}3 = 1.2490 $ , the solutions in the interval $ [0, 2) $ are 1.2490 and $ + 1.2490 = 4.3906 $ . $ t = -2 $ has solutions in quadrants II and IV; since $ ^{-1}2 = 1.1071 $ , these solutions are $ - 1.1071 = 2.0344 $ and $ 2- 1.1071 = 5.1761 $ . Extending to the entire real line, all solutions are given by $ 1.2490 + 2n $ , $ 2.0344 + 2n $ , $ 4.3906 + 2n $ , $ 5.1761 + 2n $ , n any integer. More compactly, since tangent has period $ $ , all solutions can be written as $ 1.2490 + n $ , $ 2.0344 + n $ , n any integer.

23.17. Find approximate values for all solutions on the interval $ [0, 2) $ for $ 3 ^{2} x - 5 x = 2 $ .

This is an equation in quadratic form in the quantity $ x $ . It is most efficiently solved by factoring:

\[ \begin{aligned}3\sin^{2}x-5\sin x-2&=0\ $ 3\sin x+1)(\sin x-2)&=0\\3\sin x+1&=0\quad\quad or\quad\quad\sin x-2=0\\\sin x&=\frac{-\frac{1}{3}}{}\quad\quad\sin x=2\end{aligned} \]

Use the calculator to find approximate values for solutions of these equations: $ x = - $ has solutions in quadrants III and IV; since $ ^{-1} = 0.339 $ , the solutions in the interval $ [0, 2) $ are $ + 0.3398 = 3.4814 $ and $ 2- 0.3398 = 5.9434 $ . $ x = 2 $ has no solutions, since 2 is not in the range of the sine function; a calculator will return an error message.

Solutions: 3.4814, 5.9434.

23.18. Find approximate values for all angles in the interval [0°, 360°) that satisfy 3 cos²A + 5 cos A − 1 = 0

This is an equation in quadratic form in the quantity cos A. Since it is not factorable in the integers, use the quadratic formula, with a = 3, b = 5, c = -1.

\[ \begin{aligned}\cos A&=\frac{-5\pm\sqrt{5^{2}-4(3)(-1)}}{2\cdot3}\\\cos A&=\frac{-5\pm\sqrt{37}}{6}\end{aligned} \]

Using a calculator to approximate these values yields $ A = 0.1805 $ and $ A = -1.8471 $ . The first of these has solutions in quadrants I and IV; since $ ^{-1}() = 1.3893 = 79.6^{} $ , the solutions are $ 79.6^{} $ and $ 360^{} - 79.6^{} = 280.4^{} $ . $ A = -1.8471 $ has no solutions, since -1.8471 is not in the range of the cosine function; a calculator will return an error message.

Solutions: 79.6°, 280.4°

SUPPLEMENTARY PROBLEMS

23.19. Simplify

  1. $ {2}x{2}x $

  2. cos t(1 + tan2t)

  3. $ (+ )(- ) $

  4. $ - $

Ans. (a) $ ^{2}x $ ; (b) $ t $ ; (c) -1; (d) $ 2$

23.20. Simplify (a) $ x x $ ; (b) $ 1 - $ ; (c) $ $ ; (d) $ + $

Ans. (a) $ x $ ; (b) $ x $ ; (c) $ u - u $ ; (d) $ 2x $

23.21. Verify that the following are identities:

  1. $ - x = $ (b) $ + = 2x $ (c) $ (+ )^{2} = $

23.22. Verify that the following are identities:

\[ \frac{\cos t}{\csc t-\sin t}=\tan t;(b)\sec^{2}x-(1+\tan x)^{2}=-2\tan x;(c)\tan t+\frac{1}{\tan t}=\frac{1}{\sin t\cos t} \]

23.23. Verify that the following are identities:

  1. $ x(1 - 2^{2}x + ^{4}x) = ^{5}x $ ; (b) $ = 1 + tt $ ;

  2. $ ^{2}u - ^{2}u = $

23.24. Verify that $ (x) = -(x) $ is an identity.

23.25. Simplify the following algebraic expressions by making the indicated substitution:

  1. $ $ , substitute $ x=2u $ , $ -u $ ;

  2. $ $ , substitute $ x=u $ , $ -<u< $ ;

  3. $ $ , substitute x = a sec u, a > 0, $ 0 u < $

Ans. (a) $ $ sec u csc u; (b) $ 18 sec^{2}uu $ ; (c) $ u $

23.26. Show that the following are not identities:

  1. $ = ; $ (b) $ = 2$

23.27. Find all solutions:

\[ 4\sin x+2\sqrt{3}=0 \]

  1. $ 3t = 1 $

  2. $ 2 ^{2} u = u $

  3. $ 4 - ^{2}= 1 $

  4. $ x = 0 $

Ans. (a) $ x = 4/3 + 2n $ , $ 5/3 + 2n $ ; (b) $ t = /12 + n/3 $

  1. $ u = /2 + 2n $ , $ 3/2 + 2n $ , $ /3 + 2n $ , $ 5/3 + 2n $ ; (d) no solution; (e) $ x = /2 + 2n $

23.28. Find all solutions on the interval $ [0, 2) $ .

\[ (a)2\cos^{2}4\theta=1;(b)\frac{1+\sin x}{\cos x}+\frac{\cos x}{1+\sin x}=4;(c)2\cos^{2}x+3\sin x=3;(d)\tan x-\sec x=1 \]

Ans. (a) $ = , , , , , , , ; $ (b) $ x = , ; $ (c) $ x = , , ; $ (d) $ x = $

23.29. Find approximate values for all solutions on the interval $ [0^{}, 360^{}) $ .

\[ (a)4\sin^{2}A-4\sin A-1=0;(b)2\cos^{2}2A+3\cos2A-1=0. \]

Ans. (a) 191.95°, 348.05°; (b) 36.85°, 143.15°, 216.85°, 323.15°