Trigonometric Identities and Equations

9.1 Trigonometric Identities

Trigonometric Identities

Before we solve trigonometric equations, we need a solid toolbox of identities. These identities let us rewrite expressions into forms that are easier to evaluate, simplify, or verify.

Reciprocal Identities

When you see pairs like sine/cosecant or cosine/secant, think "flip" immediately.

Definition: Reciprocal Identities

\[ \begin{align*} \csc\theta&=\frac{1}{\sin\theta} & \sec\theta&=\frac{1}{\cos\theta} & \cot\theta&=\frac{1}{\tan\theta} \end{align*} \]

Fact: Equivalent Quotient Forms

These are closely related and often used with reciprocal identities:

\[ \begin{align*} \tan\theta&=\frac{\sin\theta}{\cos\theta}, \qquad \cot\theta=\frac{\cos\theta}{\sin\theta} \end{align*} \]

Example: Simplify by rewriting with identities

Find the exact value of $$ \tan(35^\circ)-\frac{\sin(35^\circ)}{\cos(35^\circ)}. $$

Show Solution

The two terms are actually the same expression written two different ways.

\[ \begin{align*} \tan(35^\circ)-\frac{\sin(35^\circ)}{\cos(35^\circ)} &=\frac{\sin(35^\circ)}{\cos(35^\circ)}-\frac{\sin(35^\circ)}{\cos(35^\circ)}\\ &=0 \end{align*} \]

So the exact value is $0$.

Example: Simplify each expression without a calculator

  1. $\cot(x)-\frac{\cos(x)}{\sin(x)}$
  2. $\sec(x)-\frac{1}{\cos(x)}$
  3. $\sin(x)\cdot\csc(x)$
  4. $\tan(500^\circ)\cdot\cot(500^\circ)$
Show Solution

We will use reciprocal/quotient identities directly in each part.

  1. Rewrite cotangent as a quotient.

    \[ \begin{align*} \cot(x)-\frac{\cos(x)}{\sin(x)} &=\frac{\cos(x)}{\sin(x)}-\frac{\cos(x)}{\sin(x)}\\ &=0 \end{align*} \]

  2. Rewrite secant as a reciprocal.

    \[ \begin{align*} \sec(x)-\frac{1}{\cos(x)} &=\frac{1}{\cos(x)}-\frac{1}{\cos(x)}\\ &=0 \end{align*} \]

  3. Multiply a function by its reciprocal.

    \[ \begin{align*} \sin(x)\cdot\csc(x) &=\sin(x)\cdot\frac{1}{\sin(x)}\\ &=1 \end{align*} \]

  4. Multiply tangent by its reciprocal cotangent.

    \[ \begin{align*} \tan(500^\circ)\cdot\cot(500^\circ) &=\tan(500^\circ)\cdot\frac{1}{\tan(500^\circ)}\\ &=1 \end{align*} \]
MyOpenMath: Try your own!

Pythagorean Identities

These come from the unit circle and right-triangle relationships. They are often used to swap one trig function for another.

Theorem: Pythagorean Identities

\[ \begin{align*} \sin^2\theta+\cos^2\theta&=1\\ \tan^2\theta+1&=\sec^2\theta\\ 1+\cot^2\theta&=\csc^2\theta \end{align*} \]

Example: Simplify using a Pythagorean identity

Simplify the expression $$ \cos^2(\theta)+\frac{1}{\csc^2(\theta)}. $$

Show Solution

Start with the reciprocal identity:

$$ \frac{1}{\csc^2(\theta)}=\sin^2(\theta). $$

So the expression becomes

\[ \begin{align*} \cos^2(\theta)+\frac{1}{\csc^2(\theta)} &=\cos^2(\theta)+\sin^2(\theta)\\ &=1 \end{align*} \]

So the exact value is $1$.

Example: Find missing trig values from given information

Find $\tan(\theta)$ if $$ \sec(\theta)=\frac{5}{4} \qquad\text{and}\qquad \csc(\theta)=\frac{5}{3}. $$

Show Solution

Use reciprocal identities first:

$$ \cos(\theta)=\frac{4}{5}, \qquad \sin(\theta)=\frac{3}{5}. $$

Now apply the quotient identity for tangent:

\[ \begin{align*} \tan(\theta) &=\frac{\sin(\theta)}{\cos(\theta)}\\ &=\frac{\frac{3}{5}}{\frac{4}{5}}\\ &=\frac{3}{4} \end{align*} \]

So $\tan(\theta)=\frac{3}{4}$.

MyOpenMath: Try your own!

Even and Odd Identities

These are all about what happens when the input angle is negative.

Fact: Even-Odd Identities

Even functions (sign stays the same):

\[ \cos(-\theta)=\cos(\theta), \qquad \sec(-\theta)=\sec(\theta) \]

Odd functions (sign flips):

\[ \sin(-\theta)=-\sin(\theta), \quad \csc(-\theta)=-\csc(\theta), \quad \tan(-\theta)=-\tan(\theta), \quad \cot(-\theta)=-\cot(\theta) \]

Example: Evaluate using even/odd identities

Evaluate each expression exactly.

  1. $\sin(-45^\circ)$
  2. $\cos\left(-\frac{\pi}{4}\right)$
  3. $\tan\left(-\frac{2\pi}{3}\right)$
Show Solution

  1. Begin by noting that sine is odd.

    \[ \begin{align*} \sin(-45^\circ) &=-\sin(45^\circ)\\ &=-\frac{\sqrt{2}}{2} \end{align*} \]

  2. Next, use that cosine is an even function.

    \[ \begin{align*} \cos\left(-\frac{\pi}{4}\right) &=\cos\left(\frac{\pi}{4}\right)\\ &=\frac{\sqrt{2}}{2} \end{align*} \]

  3. Finally, use that tangent is odd.

    \[ \begin{align*} \tan\left(-\frac{2\pi}{3}\right) &=-\tan\left(\frac{2\pi}{3}\right)\\ &=-(-\sqrt3)\\ &=\sqrt3 \end{align*} \]

Example: Evaluate more negative-angle trig values

  1. $\sec\left(-\frac{5\pi}{6}\right)$
  2. $\csc(-45^\circ)$
  3. $\cot(-30^\circ)$
Show Solution

  1. Secant is even, so we can drop the negative inside.

    \[ \begin{align*} \sec\left(-\frac{5\pi}{6}\right) &=\sec\left(\frac{5\pi}{6}\right)\\ &=\frac{1}{\cos\left(\frac{5\pi}{6}\right)}\\ &=\frac{1}{-\frac{\sqrt3}{2}}\\ &=-\frac{2\sqrt3}{3} \end{align*} \]

  2. Cosecant is odd, so pull out a negative.

    \[ \begin{align*} \csc(-45^\circ) &=-\csc(45^\circ)\\ &=-\frac{1}{\sin(45^\circ)}\\ &=-\frac{1}{\frac{\sqrt{2}}{2}}\\ &=-\sqrt{2} \end{align*} \]

  3. Cotangent is odd, so pull out a negative.

    \[ \begin{align*} \cot(-30^\circ) &=-\cot(30^\circ)\\ &=-\frac{1}{\tan(30^\circ)}\\ &=-\frac{1}{\frac{1}{\sqrt3}}\\ &=-\sqrt3 \end{align*} \]
MyOpenMath: Try your own!

Verifying Identities

When we verify an identity, we are not solving for a variable. We are proving two expressions are equivalent.

Fact: Strategy for Verifying Identities

  1. Pick one side, usually the more complicated side.
  2. Rewrite using known identities.
  3. Use algebra carefully: factor, combine fractions, and simplify.
  4. Stop once it matches the other side exactly.

Avoid working both sides at the same time.

Example: Verify an identity

Verify: $$ \cot(\theta)\cdot\sec(\theta)=\frac{1}{\sin(\theta)}. $$

Show Solution

Start on the left-hand side.

\[ \begin{align*} \cot(\theta)\cdot\sec(\theta) &=\left(\frac{\cos\theta}{\sin\theta}\right)\left(\frac{1}{\cos\theta}\right)\\ &=\frac{1}{\sin\theta} \end{align*} \]

This matches the right-hand side, so the identity is verified.

Example: Verify an identity by rationalizing

Verify: $$ \frac{\sin\theta}{1+\cos\theta}=\frac{1-\cos\theta}{\sin\theta}. $$

Show Solution

Start with the left-hand side and multiply by the conjugate.

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

This matches the right-hand side, so the identity is verified.

MyOpenMath: Try your own!