solved problems on trigonometric identities

$Graph of y=cos(theta) from -2pi to 2pi, showing in particular that it is symmetric about the y-axis. Points given are (-pi/4, .707) and (pi/4, .707).$

For all $\theta$ in the domain of the sine and cosine functions, respectively, we can state the following:

The other even-odd identities follow from the even and odd nature of the sine and cosine functions. For example, consider the tangent identity,$\tan(−\theta)=−\tan \theta$. We can interpret the tangent of a negative angle as

\[\tan (−\theta)=\dfrac{\sin (−\theta)}{\cos (−\theta)}=\dfrac{−\sin \theta}{\cos \theta}=−\tan \theta. \nonumber\]

Tangent is therefore an odd function, which means that $\tan(−\theta)=−\tan(\theta)$ for all $\theta$ in the domain of the tangent function.

The cotangent identity, $\cot(−\theta)=−\cot \theta$,also follows from the sine and cosine identities. We can interpret the cotangent of a negative angle as

\[\cot(−\theta)=\dfrac{\cos(−\theta)}{\sin(−\theta)}=\dfrac{\cos \theta}{−\sin \theta}=−\cot \theta.\nonumber\]

Cotangent is therefore an odd function, which means that $\cot(−\theta)=−\cot(\theta)$ for all $\theta$ in the domain of the cotangent function.

The cosecant function is the reciprocal of the sine function, which means that the cosecant of a negative angle will be interpreted as

\[\csc(−\theta)=\dfrac{1}{\sin(−\theta)}=\dfrac{1}{−\sin \theta}=−\csc \theta. \nonumber\]

The cosecant function is therefore odd.

Finally, the secant function is the reciprocal of the cosine function, and the secant of a negative angle is interpreted as

\[\sec(−\theta)=\dfrac{1}{\cos(−\theta)}=\dfrac{1}{\cos \theta}=\sec \theta. \nonumber\]

The secant function is therefore even.

To sum up, only two of the trigonometric functions, cosine and secant, are even. The other four functions are odd, verifying the even-odd identities.

The next set of fundamental identities is the set of reciprocal identities, which, as their name implies, relate trigonometric functions that are reciprocals of each other. (Table $\PageIndex{3}$). Recall that we first encountered these identities when defining trigonometric functions from right angles in Right Angle Trigonometry .

The final set of identities is the set of quotient identities, which define relationships among certain trigonometric functions and can be very helpful in verifying other identities (Table $\PageIndex{4}$).

The reciprocal and quotient identities are derived from the definitions of the basic trigonometric functions.

SUMMARIZING TRIGONOMETRIC IDENTITIES

The Pythagorean identities are based on the properties of a right triangle.

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

\[1+{\cot}^2 \theta={\csc}^2 \theta\]

\[1+{\tan}^2 \theta={\sec}^2 \theta\]

The even-odd identities relate the value of a trigonometric function at a given angle to the value of the function at the opposite angle.

\[\tan(−\theta)=−\tan \theta\]

\[\cot(−\theta)=−\cot \theta\]

\[\sin(−\theta)=−\sin \theta\]

\[\csc(−\theta)=−\csc \theta\]

\[\cos(−\theta)=\cos \theta\]

\[\sec(−\theta)=\sec \theta\]

The reciprocal identities define reciprocals of the trigonometric functions.

\[\sin \theta=\dfrac{1}{\csc \theta}\]

\[\cos \theta=\dfrac{1}{\sec \theta}\]

\[\tan \theta=\dfrac{1}{\cot \theta}\]

\[\csc \theta=\dfrac{1}{\sin \theta}\]

\[\sec \theta=\dfrac{1}{\cos \theta}\]

\[\cot \theta=\dfrac{1}{\tan \theta}\]

The quotient identities define the relationship among the trigonometric functions.

\[\tan \theta=\dfrac{\sin \theta}{\cos \theta}\]

\[\cot \theta=\dfrac{\cos \theta}{\sin \theta}\]

Example $\PageIndex{1}$: Graphing the Equations of an Identity

Graph both sides of the identity $\cot \theta=\dfrac{1}{\tan \theta}$. In other words, on the graphing calculator, graph $y=\cot \theta$ and $y=\dfrac{1}{\tan \theta}$.

See Figure $\PageIndex{4}$.

$Graph of y = cot(theta) and y=1/tan(theta) from -2pi to 2pi. They are the same!$

We see only one graph because both expressions generate the same image. One is on top of the other. This is a good way to prove any identity. If both expressions give the same graph, then they must be identities.

How to: Given a trigonometric identity, verify that it is true.

Example $\PageIndex{2}$: Verifying a Trigonometric Identity

Verify $\tan \theta \cos \theta=\sin \theta$.

We will start on the left side, as it is the more complicated side:

\[ \begin{align*} \tan \theta \cos \theta &=\left(\dfrac{\sin \theta}{\cos \theta}\right)\cos \theta \\[4pt] &=\sin \theta \end{align*}\]

This identity was fairly simple to verify, as it only required writing $\tan \theta$ in terms of $\sin \theta$ and $\cos \theta$.

Exercise $\PageIndex{1}$

Verify the identity $\csc \theta \cos \theta \tan \theta=1$.

\[ \begin{align*} \csc \theta \cos \theta \tan \theta=\left(\dfrac{1}{\sin \theta}\right)\cos \theta\left(\dfrac{\sin \theta}{\cos \theta}\right) \\[4pt] & =\dfrac{\cos \theta}{\sin \theta}(\dfrac{\sin \theta}{\cos \theta}) \\[4pt] & =\dfrac{\sin \theta \cos \theta}{\sin \theta \cos \theta} \\[4pt] &=1 \end{align*}\]

Example $\PageIndex{3A}$: Verifying the Equivalency Using the Even-Odd Identities

Verify the following equivalency using the even-odd identities:

$(1+\sin x)[1+\sin(−x)]={\cos}^2 x$

Working on the left side of the equation, we have

$ (1+\sin x)[1+\sin(−x)]=(1+\sin x)(1-\sin x)$

\[\begin{align*} \sin(-x)&= -\sin x \\ [5pt] &=1-{\sin}^2 x\qquad \text{Difference of squares} \\ [5pt] &={\cos}^2 x \\ {\cos}^2 x&= 1-{\sin}^2 x \\ \end{align*}\]

Example $\PageIndex{3B}$: Verifying a Trigonometric Identity Involving ${\sec}^2 \theta$

Verify the identity $\dfrac{{\sec}^2 \theta−1}{{\sec}^2 \theta}={\sin}^2 \theta$

As the left side is more complicated, let’s begin there.

\[\begin{align*} \dfrac{{\sec}^2 \theta-1}{{\sec}^2 \theta}&= \dfrac{({\tan}^2 \theta +1)-1}{{\sec}^2 \theta}\\ {\sec}^2 \theta&= {\tan}^2 \theta +1\\ &= \dfrac{{\tan}^2 \theta}{{\sec}^2 \theta}\\ &= {\tan}^2 \theta\left (\dfrac{1}{{\sec}^2 \theta}\right )\\ &= {\tan}^2 \theta \left ({\cos}^2 \theta\right )\\ {\cos}^2 \theta&= \dfrac{1}{{\sec}^2 \theta}\\ &= \left (\dfrac{{\sin}^2 \theta}{{\cos}^2 \theta}\right )\\ {\tan}^2 \theta&= \dfrac{{\sin}^2 \theta}{{\cos}^2 \theta}\\ &= {\sin}^2 \theta \end{align*}\]

There is more than one way to verify an identity. Here is another possibility. Again, we can start with the left side.

\[\begin{align*} \dfrac{{\sec}^2 \theta-1}{{\sec}^2 \theta}&= \dfrac{{\sec}^2 \theta}{{\sec}^2 \theta}-\dfrac{1}{{\sec}^2 \theta}\\ &= 1-{\cos}^2 \theta\\ &= {\sin}^2 \theta \end{align*}\]

In the first method, we used the identity ${\sec}^2 \theta={\tan}^2 \theta+1$ and continued to simplify. In the second method, we split the fraction, putting both terms in the numerator over the common denominator. This problem illustrates that there are multiple ways we can verify an identity. Employing some creativity can sometimes simplify a procedure. As long as the substitutions are correct, the answer will be the same.

Exercise $\PageIndex{2}$

Show that $\dfrac{\cot \theta}{\csc \theta}=\cos \theta$.

\[\begin{align*} \dfrac{\cot \theta}{\csc \theta}&= \dfrac{\tfrac{\cos \theta}{\sin \theta}}{\dfrac{1}{\sin \theta}}\\ &= \dfrac{\cos \theta}{\sin \theta}\cdot \dfrac{\sin \theta}{1}\\ &= \cos \theta \end{align*}\]

Example $\PageIndex{4}$: Creating and Verifying an Identity

Create an identity for the expression $2 \tan \theta \sec \theta$ by rewriting strictly in terms of sine.

There are a number of ways to begin, but here we will use the quotient and reciprocal identities to rewrite the expression:

\[\begin{align*} 2 \tan \theta \sec \theta&= 2\left (\dfrac{\sin \theta}{\cos \theta}\right )\left(\dfrac{1}{\cos \theta}\right )\\ &= \dfrac{2\sin \theta}{{\cos}^2 \theta}\\ &= \dfrac{2\sin \theta}{1-{\sin}^2 \theta}\qquad \text{Substitute } 1-{\sin}^2 \theta \text{ for } {\cos}^2 \theta \end{align*}\]

$2 \tan \theta \sec \theta=\dfrac{2 \sin \theta}{1−{\sin}^2 \theta}$

Example $\PageIndex{5}$: Verifying an Identity Using Algebra and Even/Odd Identities

Verify the identity:

$\dfrac{{\sin}^2(−\theta)−{\cos}^2(−\theta)}{\sin(−\theta)−\cos(−\theta)}=\cos \theta−\sin \theta$

Let’s start with the left side and simplify:

\[\begin{align*} \dfrac{{\sin}^2(-\theta)-{\cos}^2(-\theta)}{\sin(-\theta)-\cos(-\theta)}&= \dfrac{{[\sin(-\theta)]}^2-{[\cos(-\theta)]}^2}{\sin(-\theta)-\cos(-\theta)}\\ &= \dfrac{{(-\sin \theta)}^2-{(\cos \theta)}^2}{-\sin \theta -\cos \theta} \;\; \; , \sin(-x) = -\sin\space x\text { and } \cos(-x)=\cos \space x\\ &= \dfrac{{(\sin \theta)}^2-{(\cos \theta)}^2}{-\sin \theta -\cos \theta}\qquad \text{Difference of squares}\\ &= \dfrac{(\sin \theta-\cos \theta)(\sin \theta+\cos \theta)}{-(\sin \theta+\cos \theta)}\\ &= \cos \theta-\sin \theta \end{align*}\]

Exercise $\PageIndex{3}$

Verify the identity $\dfrac{{\sin}^2 \theta−1}{\tan \theta \sin \theta−\tan \theta}=\dfrac{\sin \theta+1}{\tan \theta}$.

\[\begin{align*} \dfrac{{\sin}^2 \theta-1}{\tan \theta \sin \theta-\tan \theta}&= \dfrac{(\sin \theta +1)(\sin \theta -1)}{\tan \theta(\sin \theta -1)}\\ &= \dfrac{\sin \theta+1}{\tan \theta} \end{align*}\]

Example $\PageIndex{6}$: Verifying an Identity Involving Cosines and Cotangents

Verify the identity: $(1−{\cos}^2 x)(1+{\cot}^2 x)=1$.

\[\begin{align*} (1-{\cos}^2 x)(1+{\cot}^2 x)&= (1-{\cos}^2 x)\left(1+\dfrac{{\cos}^2 x}{{\sin}^2 x}\right)\\ &= (1-{\cos}^2 x)\left(\dfrac{{\sin}^2 x}{{\sin}^2 x}+\dfrac{{\cos}^2 x}{{\sin}^2 x}\right )\qquad \text{Find the common denominator}\\ &= (1-{\cos}^2 x)\left(\dfrac{{\sin}^2 x +{\cos}^2 x}{{\sin}^2 x}\right)\\ &= ({\sin}^2 x)\left (\dfrac{1}{{\sin}^2 x}\right )\\ &= 1 \end{align*}\]

Using Algebra to Simplify Trigonometric Expressions

For example, the equation $(\sin x+1)(\sin x−1)=0$ resembles the equation $(x+1)(x−1)=0$, which uses the factored form of the difference of squares. Using algebra makes finding a solution straightforward and familiar. We can set each factor equal to zero and solve. This is one example of recognizing algebraic patterns in trigonometric expressions or equations.

Another example is the difference of squares formula, $a^2−b^2=(a−b)(a+b)$, which is widely used in many areas other than mathematics, such as engineering, architecture, and physics. We can also create our own identities by continually expanding an expression and making the appropriate substitutions. Using algebraic properties and formulas makes many trigonometric equations easier to understand and solve.

Example $\PageIndex{7A}$: Writing the Trigonometric Expression as an Algebraic Expression

Write the following trigonometric expression as an algebraic expression: $2{\cos}^2 \theta+\cos \theta−1$.

Notice that the pattern displayed has the same form as a standard quadratic expression,$ax^2+bx+c$. Letting $\cos \theta=x$,we can rewrite the expression as follows:

$2x^2+x−1$

This expression can be factored as $(2x+1)(x−1)$. If it were set equal to zero and we wanted to solve the equation, we would use the zero factor property and solve each factor for $x$. At this point, we would replace $x$ with $\cos \theta$ and solve for $\theta$.

Example $\PageIndex{7B}$: Rewriting a Trigonometric Expression Using the Difference of Squares

Rewrite the trigonometric expression using the difference of squares: $4{cos}^2 \theta−1$.

Notice that both the coefficient and the trigonometric expression in the first term are squared, and the square of the number 1 is 1. This is the difference of squares.

\[\begin{align*} 4{\cos}^2 \theta-1&= {(2\cos \theta)}^2-1\\ &= (2\cos \theta-1)(2\cos \theta+1) \end{align*}\]

If this expression were written in the form of an equation set equal to zero, we could solve each factor using the zero factor property. We could also use substitution like we did in the previous problem and let $\cos \theta=x$, rewrite the expression as $4x^2−1$, and factor $(2x−1)(2x+1)$. Then replace $x$ with $\cos \theta$ and solve for the angle.

Exercise $\PageIndex{4}$

Rewrite the trigonometric expression using the difference of squares: $25−9{\sin}^2 \theta$.

This is a difference of squares formula: $25−9{\sin}^2 \theta=(5−3\sin \theta)(5+3\sin \theta)$.

Example $\PageIndex{8}$: Simplify by Rewriting and Using Substitution

Simplify the expression by rewriting and using identities:

${\csc}^2 \theta−{\cot}^2 \theta$

We can start with the Pythagorean identity.

\[\begin{align*} 1+{\cot}^2 \theta&= {\csc}^2 \theta\\ \text{Now we can simplify by substituting } 1+{\cot}^2 \theta \text{ for } {\csc}^2 \theta\\ {\csc}^2 \theta-{\cot}^2 \theta&= 1+{\cot}^2 \theta-{\cot}^2 \theta\\ &= 1 \end{align*}\]

Exercise $\PageIndex{5}$

Use algebraic techniques to verify the identity: $\dfrac{\cos \theta}{1+\sin \theta}=\dfrac{1−\sin \theta}{\cos \theta}$.

(Hint: Multiply the numerator and denominator on the left side by $1−\sin \theta$.)

\[\begin{align*} \dfrac{\cos \theta}{1+\sin \theta}\left(\dfrac{1-\sin \theta}{1-\sin \theta}\right)&= \dfrac{\cos \theta (1-\sin \theta)}{1-{\sin}^2 \theta}\\ &= \dfrac{\cos \theta (1-\sin \theta)}{{\cos}^2 \theta}\\ &= \dfrac{1-\sin \theta}{\cos \theta} \end{align*}\]

Access these online resources for additional instruction and practice with the fundamental trigonometric identities.

Key Equations

Key concepts.

IIT JEE Study Material

Trigonometric Equations and Identities Solved Examples

Trigonometric Equations and Identities Solved Examples are given on this page. Trigonometric identities have different formulae, which are inequalities involving trigonometric functions of one or more angles. Trigonometric equations and identities are useful to solve trigonometric problems.

Consider a right-angled triangle ABC. Here ∠CAB = A and ∠BCA = 90°. AB is called hypotenuse of a triangle. The trigonometric ratios are defined using the adjacent, opposite and hypotenuse sides of the triangle .

BC / AB = (Opposite side) / hypotenuse, gives sine of A and denoted as sin A.

AC / AB = (adjacent side) / hypotenuse, gives the cosine of A and denoted as cos A.

BC / AC = (Opposite side) /(adjacent side), gives the tangent of A and denoted as tan A.

cosec A = 1/sin A

sec A = 1/cos A

cot A = 1/tan A

Trigonometric Equations And Identities

An equation which consists of 1 or more trigonometric ratios of angles that are unknown is called a trigonometric equation. A trigonometric equation is written as P1 (sin θ, cos θ, tan θ, cot θ, sec θ, cosec θ) = P2 (sin θ, cos θ, tan θ, cot θ, sec θ, cosec θ), where P1 and P2 are rational functions.

A trigonometric equation which is always true and holds good for every angle is called a trigonometric identity.

Steps to solve a trigonometric equation are as follows:

Type – I

The first type of equations involves factorising them or expressing in quadratic form. They can be solved by converting them into factors and find the solution for each factor. The solution obtained finally is the association of solutions of all the factors.

This type of equation can be written in the form of cos ⁡A + b sin⁡ B = c and then find the solution.

List of Trigonometric Equations And Identities

Trigonometric Identities

a) sin 2 θ + cos 2 θ = 1

b) sec 2 θ = 1 + tan 2 θ

c) cosec 2 θ = 1 + cot 2 θ

General solution of some Trigonometric equations

a) sin θ = 0 ⇒ θ = n π

b) tan θ = 0 ⇒ θ = n π

c) cos θ = 0 ⇒ θ = (2n + 1) π/2

d) sin θ = 1 ⇒ θ = (4n + 1) π/2

e) cos θ = – 1 ⇒ θ = (4n – 1) π/2

f) cos θ = 1 ⇒ θ = 2nπ

g) cos = – 1 ⇒ θ = (2n + 1) π

h) cot θ = 0 ⇒ θ (2n + 1) π/2

General Solution of some Standard Equations

a) sin θ = sin α ⇒ θ = nπ + (-1) n θ

b) cos θ = cos α ⇒ θ = 2nπ ± θ

c) tan θ = tan α ⇒ θ = nπ +θ

Angle-Sum and Difference Identities

Negative Angle identities

Trigonometric Equations and its Solutions

Trigonometry Previous Year Questions with Solutions

Trigonometric Equations And Identities Examples

Some Trigonometric identities problems are given below. JEE aspirants are advised to learn Trigonometric Equations And Identities Solved Examples so that they can score better ranks.

Example 1: In Δ ABC, sin (A − B) / sin (A + B) = ?

But cos B = [a 2 + c 2 − b 2 ] / 2ac,cos A = b 2 + c 2 − a 2 / 2bc

⇒ [a / c] [cos B] − [b / c] [cos A] = 1 / 2c 2

= (a 2 + c 2 − b 2 − b 2 – c 2 + a 2 ) / 2c 2

Example 2: If the angles of a triangle are in the ratio 1: 2: 7, then what is the ratio of its greatest side to the least side?

x + 2x + 7x = 180 o

Hence, the angles are 18 o , 36 o , 126 o

Greatest side ∝ sin(126 o ) Smallest side ∝ sin(18 o ) and

Ratio = sin126 o / sin(18 o )

Example 3: The value of sin 2 5 o + sin 2 10 o + sin 2 15 o + . . . + sin 2 85 o + sin 2 90 o is equal to

Given expression is sin 2 5 o + sin 2 10 o + sin 2 15 o + . . . + sin 2 85 o + sin 2 90 o .

We know that sin 90 o = 1 or sin 2 90 o = 1.

We also know that sin 2 85 o = [sin (90 o −5 o )] 2 = cos 2 5 o .

Therefore, from the complementary rule, we find sin 2 5 o + sin 2 85 o = sin 2 5 o +cos 2 5 o = 1

Therefore, sin 2 5 o + sin 2 10 o + sin 2 15 o + . . . + sin 2 85 o + sin 2 90 o =(1 + 1 + 1+ 1 + 1 + 1 + 1 + 1 ) + 1 + 1 / 2 = 9 [1 / 2].

Example 4: Find the value of 1 + cos 56 o + cos58 o − cos66 o .

1 + cos 56 o + cos58 o − cos66 o

Apply the conditional identity

We get the value of the required expression equal to 4 cos28 o cos29 o sin33 o .

Example 5: If a cos 3 α + 3a cos α sin 2 α = m and a sin 3 α + 3a cos 2 α sin α = n, then find (m + n) 2/3 + (m – n) 2/3 .

Adding and subtracting the given relation, we get

$\begin{array}{l}(m+n)=a{{\cos }^{3}}\alpha +3a\cos \alpha \,{{\sin }^{2}}\alpha +3a{{\cos }^{2}}\alpha .\sin \alpha +a{{\sin }^{3}}\alpha =a{{(\cos \alpha +\sin \alpha )}^{3}}\end{array} $ and similarly

$\begin{array}{l}(m-n)=a\,\,{{(\cos \alpha -\sin \alpha )}^{3}}\end{array} $

$\begin{array}{l}{{(m+n)}^{2/3}}+{{(m-n)}^{2/3}}={{a}^{2/3}}{{\{\cos \alpha +\sin \alpha )}^{2}}+{{(\cos \alpha -\sin \alpha )}^{2}}\}\\={{a}^{2/3}}\{2({{\cos }^{2}}\alpha +{{\sin }^{2}}\alpha )\}=2{{a}^{2/3}}\end{array} $

Example 6: If sin A = n sin B, then $\begin{array}{l}\frac{n-1}{n+1}\tan \,\frac{A+B}{2}=\end{array} $

$\begin{array}{l}\sin A=n\sin B\Rightarrow \frac{n}{1}=\frac{\sin A}{\sin B}\\ \Rightarrow \frac{n-1}{n+1}=\frac{\sin A-\sin B}{\sin A+\sin B}=\frac{2\cos \frac{A+B}{2}\sin \frac{A-B}{2}}{2\sin \frac{A+B}{2}\cos \frac{A-B}{2}}\\ =\tan \frac{A-B}{2}\cot \frac{A+B}{2}\\ \Rightarrow \frac{n-1}{n+1}\tan \left( \frac{A+B}{2} \right)\\ =\tan \frac{A-B}{2}.\end{array} $

Example 7: If $\begin{array}{l}\tan \theta =\frac{\sin \alpha -\cos \alpha }{\sin \alpha +\cos \alpha },\ \text{then}\end{array} $ sin α + cos α and sin α – cos α.

$\begin{array}{l}\tan \theta =\frac{\sin \alpha -\cos \alpha }{\sin \alpha +\cos \alpha } \\ \Rightarrow \tan \theta =\frac{\sin \left( \alpha -\frac{\pi }{4} \right)}{\cos \left( \alpha -\frac{\pi }{4} \right)}\Rightarrow \tan \theta =\tan \left( \alpha -\frac{\pi }{4} \right)\\ \Rightarrow \theta =\alpha -\frac{\pi }{4}\Rightarrow \alpha =\theta +\frac{\pi }{4}\\ \sin \alpha +\cos \alpha =\sin \left( \theta +\frac{\pi }{4} \right)+\cos \left( \theta +\frac{\pi }{4} \right)\\ =\sqrt{2}\cos \theta \text \ and \ \sin \alpha -\cos \alpha =\sin \left( \theta +\frac{\pi }{4} \right)-\cos \left( \theta +\frac{\pi }{4} \right)\\ =\frac{1}{\sqrt{2}}\sin \theta +\frac{1}{\sqrt{2}}\cos \theta -\frac{1}{\sqrt{2}}\cos \theta +\frac{1}{\sqrt{2}}\sin \theta \\ =\frac{2}{\sqrt{2}}\sin \theta =\sqrt{2}\sin \theta\end{array} $

Example 8: Evaluate tan π/8.

Let x = π/8

We have tan 2x = 2 tan x /(1-tan 2 x)

So tan π/4 = 2 tan π/8 /(1-tan 2 π/8)

π/8 lies in the first quadrant. So tan π/8 is positive.

Hence tan π/8 = √2 -1.

Trigonometric Equations – Important Topics

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Module 9: Trigonometric Identities and Equations

Solving trigonometric equations with identities, learning outcomes.

Verify the fundamental trigonometric identities

To verify the trigonometric identities, we usually start with the more complicated side of the equation and essentially rewrite the expression until it has been transformed into the same expression as the other side of the equation. Sometimes we have to factor expressions, expand expressions, find common denominators, or use other algebraic strategies to obtain the desired result. In this first section, we will work with the fundamental identities: the Pythagorean identities , the even-odd identities, the reciprocal identities, and the quotient identities.

We will begin with the Pythagorean identities , which are equations involving trigonometric functions based on the properties of a right triangle. We have already seen and used the first of these identifies, but now we will also use additional identities.

The second and third identities can be obtained by manipulating the first. The identity [latex]1+{\cot }^{2}\theta ={\csc }^{2}\theta\[/latex] is found by rewriting the left side of the equation in terms of sine and cosine.

Prove: [latex]1+{\cot }^{2}\theta ={\csc }^{2}\theta [/latex]

The next set of fundamental identities is the set of even-odd identities . The even-odd identities relate the value of a trigonometric function at a given angle to the value of the function at the opposite angle and determine whether the identity is odd or even.

Recall that an odd function is one in which [latex]f\left(-x\right)= -f\left(x\right)[/latex] for all [latex]x[/latex] in the domain of [latex]f[/latex]. The sine function is an odd function because [latex]\sin \left(-\theta \right)=-\sin \theta[/latex]. The graph of an odd function is symmetric about the origin. For example, consider corresponding inputs of [latex]\frac{\pi }{2}\\[/latex] and [latex]-\frac{\pi }{2}[/latex]. The output of [latex]\sin \left(\frac{\pi }{2}\right)[/latex] is opposite the output of [latex]\sin \left(-\frac{\pi }{2}\right)[/latex]. Thus,

This is shown in Figure 2.

Figure 2. Graph of [latex]y=\sin \theta[/latex]

Recall that an even function is one in which

The graph of an even function is symmetric about the y- axis. The cosine function is an even function because [latex]\cos \left(-\theta \right)=\cos \theta[/latex]. For example, consider corresponding inputs [latex]\frac{\pi }{4}[/latex] and [latex]-\frac{\pi }{4}[/latex]. The output of [latex]\cos \left(\frac{\pi }{4}\right)[/latex] is the same as the output of [latex]\cos \left(-\frac{\pi }{4}\right)[/latex]. Thus,

Figure 3. Graph of [latex]y=\cos \theta[/latex]

For all [latex]\theta[/latex] in the domain of the sine and cosine functions, respectively, we can state the following:

The other even-odd identities follow from the even and odd nature of the sine and cosine functions. For example, consider the tangent identity, [latex]\tan \left(-\theta \right)=\mathrm{-tan}\theta[/latex]. We can interpret the tangent of a negative angle as [latex]\tan \left(-\theta \right)=\frac{\sin \left(-\theta \right)}{\cos \left(-\theta \right)}=\frac{-\sin \theta }{\cos \theta }=-\tan \theta[/latex]. Tangent is therefore an odd function, which means that [latex]\tan \left(-\theta \right)=-\tan \left(\theta \right)[/latex] for all [latex]\theta[/latex] in the domain of the tangent function .

The cotangent identity, [latex]\cot \left(-\theta \right)=-\cot \theta[/latex], also follows from the sine and cosine identities. We can interpret the cotangent of a negative angle as [latex]\cot \left(-\theta \right)=\frac{\cos \left(-\theta \right)}{\sin \left(-\theta \right)}=\frac{\cos \theta }{-\sin \theta }=-\cot \theta[/latex]. Cotangent is therefore an odd function, which means that [latex]\cot \left(-\theta \right)=-\cot \left(\theta \right)[/latex] for all [latex]\theta[/latex] in the domain of the cotangent function .

The cosecant function is the reciprocal of the sine function, which means that the cosecant of a negative angle will be interpreted as [latex]\csc \left(-\theta \right)=\frac{1}{\sin \left(-\theta \right)}=\frac{1}{-\sin \theta }=-\csc \theta[/latex]. The cosecant function is therefore odd.

Finally, the secant function is the reciprocal of the cosine function, and the secant of a negative angle is interpreted as [latex]\sec \left(-\theta \right)=\frac{1}{\cos \left(-\theta \right)}=\frac{1}{\cos \theta }=\sec \theta[/latex]. The secant function is therefore even.

To sum up, only two of the trigonometric functions, cosine and secant, are even. The other four functions are odd, verifying the even-odd identities.

The next set of fundamental identities is the set of reciprocal identities , which, as their name implies, relate trigonometric functions that are reciprocals of each other.

The final set of identities is the set of quotient identities , which define relationships among certain trigonometric functions and can be very helpful in verifying other identities.

The reciprocal and quotient identities are derived from the definitions of the basic trigonometric functions.

A General Note: Summarizing Trigonometric Identities

The Pythagorean identities are based on the properties of a right triangle.

[latex]\begin{gathered} {\cos}^{2}\theta + {\sin}^{2}\theta=1 \\ 1+{\tan}^{2}\theta={\sec}^{2}\theta \\ 1+{\cot}^{2}\theta={\csc}^{2}\theta\end{gathered}[/latex]

The even-odd identities relate the value of a trigonometric function at a given angle to the value of the function at the opposite angle.

[latex]\begin{gathered} \cos(-\theta)=\cos(\theta) \\\sin(-\theta)=-\sin(\theta) \\\tan(-\theta)=-\tan(\theta) \\\cot(-\theta)=-\cot(\theta) \\\sec(-\theta)=\sec(\theta) \\\csc(-\theta)=-\csc(\theta) \end{gathered}[/latex]

The reciprocal identities define reciprocals of the trigonometric functions.

[latex]\begin{gathered}\sin\theta=\frac{1}{\csc\theta} \\ \cos\theta=\frac{1}{\sec\theta} \\ \tan\theta=\frac{1}{\cot\theta} \\ \cot\theta=\frac{1}{\tan\theta} \\ \sec\theta=\frac{1}{\cos\theta} \\ \csc\theta=\frac{1}{\sin\theta}\end{gathered}[/latex]

The quotient identities define the relationship among the trigonometric functions.

[latex]\begin{gathered} \tan\theta=\frac{\sin\theta}{\cos\theta} \\ \cot\theta=\frac{\cos\theta}{\sin\theta} \end{gathered}[/latex]

Example 1: Graphing the Equations of an Identity

Graph both sides of the identity [latex]\cot \theta =\frac{1}{\tan \theta }[/latex]. In other words, on the graphing calculator, graph [latex]y=\cot \theta[/latex] and [latex]y=\frac{1}{\tan \theta }[/latex].

Analysis of the Solution

How To: Given a trigonometric identity, verify that it is true.

Example 2: Verifying a Trigonometric Identity

Verify [latex]\tan \theta \cos \theta =\sin \theta[/latex].

We will start on the left side, as it is the more complicated side:

[latex]\begin{align}\tan \theta \cos \theta &=\left(\frac{\sin \theta }{\cos \theta }\right)\cos \theta \\ &=\left(\frac{\sin \theta }{\cancel{\cos \theta }}\right)\cancel{\cos \theta } \\ &=\sin \theta \end{align}[/latex]

This identity was fairly simple to verify, as it only required writing [latex]\tan \theta[/latex] in terms of [latex]\sin \theta[/latex] and [latex]\cos \theta[/latex].

Verify the identity [latex]\csc \theta \cos \theta \tan \theta =1[/latex].

[latex]\begin{align}\csc \theta \cos \theta \tan \theta &=\left(\frac{1}{\sin \theta }\right)\cos \theta \left(\frac{\sin \theta }{\cos \theta }\right) \\ &=\frac{\cos \theta }{\sin \theta }\left(\frac{\sin \theta }{\cos \theta }\right) \\ &=\frac{\sin \theta \cos \theta }{\sin \theta \cos \theta } \\ &=1\end{align}[/latex]

Example 3: Verifying the Equivalency Using the Even-Odd Identities

Verify the following equivalency using the even-odd identities:

[latex]\left(1+\sin x\right)\left[1+\sin \left(-x\right)\right]={\cos }^{2}x[/latex]

Working on the left side of the equation, we have

[latex]\begin{align}\left(1+\sin x\right)\left[1+\sin \left(-x\right)\right]&=\left(1+\sin x\right)\left(1-\sin x\right)&& \text{Since sin(-}x\text{)=}-\sin x \\ &=1-{\sin }^{2}x&& \text{Difference of squares} \\ &={\cos }^{2}x&& {\text{cos}}^{2}x=1-{\sin }^{2}x\end{align}[/latex]

Example 4: Verifying a Trigonometric Identity Involving sec 2 θ

Verify the identity [latex]\frac{{\sec }^{2}\theta -1}{{\sec }^{2}\theta }={\sin }^{2}\theta[/latex]

As the left side is more complicated, let’s begin there.

[latex]\begin{align}\frac{{\sec }^{2}\theta -1}{{\sec }^{2}\theta }&=\frac{\left({\tan }^{2}\theta +1\right)-1}{{\sec }^{2}\theta }&& {\sec}^{2}\theta ={\tan }^{2}\theta +1 \\ &=\frac{{\tan }^{2}\theta }{{\sec }^{2}\theta } \\ &={\tan }^{2}\theta \left(\frac{1}{{\sec }^{2}\theta }\right) \\ &={\tan }^{2}\theta \left({\cos }^{2}\theta \right)&& {\cos }^{2}\theta =\frac{1}{{\sec }^{2}\theta } \\ &=\left(\frac{{\sin }^{2}\theta }{{\cos }^{2}\theta }\right)\left({\cos }^{2}\theta \right)&& {\tan}^{2}\theta =\frac{{\sin }^{2}\theta }{{\cos }^{2}\theta } \\ &=\left(\frac{{\sin }^{2}\theta }{\cancel{{\cos }^{2}\theta}}\right)\left(\cancel{{\cos }^{2}\theta} \right) \\ &={\sin }^{2}\theta \end{align}[/latex]

There is more than one way to verify an identity. Here is another possibility. Again, we can start with the left side.

[latex]\begin{align}\frac{{\sec }^{2}\theta -1}{{\sec }^{2}\theta }&=\frac{{\sec }^{2}\theta }{{\sec }^{2}\theta }-\frac{1}{{\sec }^{2}\theta } \\ &=1-{\cos }^{2}\theta \\ &={\sin }^{2}\theta \end{align}[/latex]

In the first method, we used the identity [latex]{\sec }^{2}\theta ={\tan }^{2}\theta +1\\[/latex] and continued to simplify. In the second method, we split the fraction, putting both terms in the numerator over the common denominator. This problem illustrates that there are multiple ways we can verify an identity. Employing some creativity can sometimes simplify a procedure. As long as the substitutions are correct, the answer will be the same.

Show that [latex]\frac{\cot \theta }{\csc \theta }=\cos \theta[/latex].

[latex]\begin{align}\frac{\cot \theta }{\csc \theta }&=\frac{\frac{\cos \theta }{\sin \theta }}{\frac{1}{\sin \theta }} \\ &=\frac{\cos \theta }{\sin \theta }\cdot \frac{\sin \theta }{1} \\ &=\cos \theta \end{align}[/latex]

Example 5: Creating and Verifying an Identity

Create an identity for the expression [latex]2\tan \theta \sec \theta[/latex] by rewriting strictly in terms of sine.

There are a number of ways to begin, but here we will use the quotient and reciprocal identities to rewrite the expression:

[latex]\begin{align}2\tan \theta \sec \theta &=2\left(\frac{\sin \theta }{\cos \theta }\right)\left(\frac{1}{\cos \theta }\right) \\ &=\frac{2\sin \theta }{{\cos }^{2}\theta } \\ &=\frac{2\sin \theta }{1-{\sin }^{2}\theta }&& \text{Substitute }1-{\sin }^{2}\theta \text{ for }{\cos }^{2}\theta \end{align}[/latex]

[latex]2\tan \theta \sec \theta =\frac{2\sin \theta }{1-{\sin }^{2}\theta }[/latex]

Example 6: Verifying an Identity Using Algebra and Even/Odd Identities

Verify the identity:

[latex]\begin{align}\frac{{\sin }^{2}\left(-\theta \right)-{\cos }^{2}\left(-\theta \right)}{\sin \left(-\theta \right)-\cos \left(-\theta \right)}=\cos \theta -\sin \theta\end{align}[/latex]

Let’s start with the left side and simplify:

Verify the identity [latex]\frac{{\sin }^{2}\theta -1}{\tan \theta \sin \theta -\tan \theta }=\frac{\sin \theta +1}{\tan \theta }[/latex].

[latex]\begin{align}\frac{{\sin }^{2}\theta -1}{\tan \theta \sin \theta -\tan \theta }&=\frac{\left(\sin \theta +1\right)\left(\sin \theta -1\right)}{\tan \theta \left(\sin \theta -1\right)}\\ &=\frac{\sin \theta +1}{\tan \theta }\end{align}[/latex]

Example 7: Verifying an Identity Involving Cosines and Cotangents

Verify the identity: [latex]\left(1-{\cos }^{2}x\right)\left(1+{\cot }^{2}x\right)=1[/latex].

We will work on the left side of the equation.

[latex]\begin{align}\left(1-{\cos }^{2}x\right)\left(1+{\cot }^{2}x\right)&=\left(1-{\cos }^{2}x\right)\left(1+\frac{{\cos }^{2}x}{{\sin }^{2}x}\right) \\ &=\left(1-{\cos }^{2}x\right)\left(\frac{{\sin }^{2}x}{{\sin }^{2}x}+\frac{{\cos }^{2}x}{{\sin }^{2}x}\right) && \text{Find the common denominator}. \\ &=\left(1-{\cos }^{2}x\right)\left(\frac{{\sin }^{2}x+{\cos }^{2}x}{{\sin }^{2}x}\right) \\ &=\left({\sin }^{2}x\right)\left(\frac{1}{{\sin }^{2}x}\right) \\ &=1\end{align}[/latex]

Simplify trigonometric expressions using algebra and the identities

For example, the equation [latex]\left(\sin x+1\right)\left(\sin x - 1\right)=0[/latex] resembles the equation [latex]\left(x+1\right)\left(x - 1\right)=0[/latex], which uses the factored form of the difference of squares. Using algebra makes finding a solution straightforward and familiar. We can set each factor equal to zero and solve. This is one example of recognizing algebraic patterns in trigonometric expressions or equations.

Another example is the difference of squares formula, [latex]{a}^{2}-{b}^{2}=\left(a-b\right)\left(a+b\right)[/latex], which is widely used in many areas other than mathematics, such as engineering, architecture, and physics. We can also create our own identities by continually expanding an expression and making the appropriate substitutions. Using algebraic properties and formulas makes many trigonometric equations easier to understand and solve.

Example 8: Writing the Trigonometric Expression as an Algebraic Expression

Write the following trigonometric expression as an algebraic expression: [latex]2{\cos }^{2}\theta +\cos \theta -1[/latex].

Notice that the pattern displayed has the same form as a standard quadratic expression, [latex]a{x}^{2}+bx+c[/latex]. Letting [latex]\cos \theta =x[/latex], we can rewrite the expression as follows:

[latex]2{x}^{2}+x - 1[/latex]

This expression can be factored as [latex]\left(2x+1\right)\left(x - 1\right)[/latex]. If it were set equal to zero and we wanted to solve the equation, we would use the zero factor property and solve each factor for [latex]x[/latex]. At this point, we would replace [latex]x[/latex] with [latex]\cos \theta [/latex] and solve for [latex]\theta [/latex].

Example 9: Rewriting a Trigonometric Expression Using the Difference of Squares

Rewrite the trigonometric expression: [latex]4{\cos }^{2}\theta -1[/latex].

Notice that both the coefficient and the trigonometric expression in the first term are squared, and the square of the number 1 is 1. This is the difference of squares. Thus,

[latex]\begin{align}4{\cos }^{2}\theta -1&={\left(2\cos \theta \right)}^{2}-1 \\ &=\left(2\cos \theta -1\right)\left(2\cos \theta +1\right) \end{align}[/latex]

If this expression were written in the form of an equation set equal to zero, we could solve each factor using the zero factor property. We could also use substitution like we did in the previous problem and let [latex]\cos \theta =x[/latex], rewrite the expression as [latex]4{x}^{2}-1[/latex], and factor [latex]\left(2x - 1\right)\left(2x+1\right)[/latex]. Then replace [latex]x[/latex] with [latex]\cos \theta [/latex] and solve for the angle.

Rewrite the trigonometric expression: [latex]25 - 9{\sin }^{2}\theta [/latex].

This is a difference of squares formula: [latex]25 - 9{\sin }^{2}\theta =\left(5 - 3\sin \theta \right)\left(5+3\sin \theta \right)[/latex].

Example 10: Simplify by Rewriting and Using Substitution

Simplify the expression by rewriting and using identities:

[latex]{\csc }^{2}\theta -{\cot }^{2}\theta [/latex]

We can start with the Pythagorean identity.

[latex]1+{\cot }^{2}\theta ={\csc }^{2}\theta [/latex]

Now we can simplify by substituting [latex]1+{\cot }^{2}\theta [/latex] for [latex]{\csc }^{2}\theta [/latex]. We have

[latex]\begin{align}{\csc }^{2}\theta -{\cot }^{2}\theta &=1+{\cot }^{2}\theta -{\cot }^{2}\theta \\ &=1\end{align}[/latex]

Use algebraic techniques to verify the identity: [latex]\frac{\cos\theta}{1+\sin\theta}=\frac{1-\sin\theta}{\cos\theta}[/latex].

(Hint: Multiply the numerator and denominator on the left side by [latex]1-\sin\theta[/latex]).

[latex]\begin{align}\frac{\cos \theta }{1+\sin \theta }\left(\frac{1-\sin \theta }{1-\sin \theta }\right)&=\frac{\cos \theta \left(1-\sin \theta \right)}{1-{\sin }^{2}\theta } \\ &=\frac{\cos \theta \left(1-\sin \theta \right)}{{\cos }^{2}\theta } \\ &=\frac{1-\sin \theta }{\cos \theta } \end{align}[/latex]

Key Equations

Key concepts.

even-odd identities

Pythagorean identities

quotient identities

reciprocal identities

Precalculus. Authored by : OpenStax College. Provided by : OpenStax. Located at : http://cnx.org/contents/f[email protected]:1/Preface . License : CC BY: Attribution

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