Proving identities
An identity is an equation that is true for every value of the variable where both sides are defined. For example, ${\sin^2 x+\cos^2 x=1}$ is an identity. The equation ${\sin x=\dfrac{1}{2}}$ is not: it is true only for some values of ${x}$.
This page collects the basic identities, uses them to simplify expressions, and shows how to prove a new identity.
The fundamental identities
These come from the definitions in Chapter 1. Learn them by heart; every proof uses them.
Reciprocal identities
${\csc x=\dfrac{1}{\sin x}}$${\sec x=\dfrac{1}{\cos x}}$${\cot x=\dfrac{1}{\tan x}}$
Quotient identities
${\tan x=\dfrac{\sin x}{\cos x}}$${\cot x=\dfrac{\cos x}{\sin x}}$
Pythagorean identities
${\sin^2 x+\cos^2 x=1}$
${1+\tan^2 x=\sec^2 x}$
${1+\cot^2 x=\csc^2 x}$
Even-odd identities
${\sin(-x)=-\sin x}$${\cos(-x)=\cos x}$${\tan(-x)=-\tan x}$
Cofunction identities
${\sin\left(\dfrac{\pi}{2}-x\right)=\cos x}$${\cos\left(\dfrac{\pi}{2}-x\right)=\sin x}$
${\tan\left(\dfrac{\pi}{2}-x\right)=\cot x}$
Each Pythagorean identity can be rearranged. For example, ${\sin^2 x=1-\cos^2 x}$ and ${\sec^2 x-\tan^2 x=1}$. Watch for these forms too.
Simplifying expressions
A good first step is to write everything in terms of sine and cosine. Then use algebra: combine fractions, factor, and cancel.
Example 1: Simplify ${\cos t+\tan t\,\sin t}$.
Solution:
Write ${\tan t}$ as ${\dfrac{\sin t}{\cos t}}$, then use the common denominator ${\cos t}$:
$\begin{align*}&\cos t+\dfrac{\sin t}{\cos t}\cdot\sin t\\&=\dfrac{\cos^2 t}{\cos t}+\dfrac{\sin^2 t}{\cos t}\\&=\dfrac{\cos^2 t+\sin^2 t}{\cos t}\\&=\dfrac{1}{\cos t}=\sec t\end{align*}$
Example 2: Simplify ${\dfrac{1-\cos^2 x}{\sin x\cos x}}$.
Solution:
By the Pythagorean identity, ${1-\cos^2 x=\sin^2 x}$:
$\begin{align*}&\dfrac{1-\cos^2 x}{\sin x\cos x}\\&=\dfrac{\sin^2 x}{\sin x\cos x}\\&=\dfrac{\sin x}{\cos x}=\tan x\end{align*}$
Is it an identity?
A graph can tell you whether a claimed identity might be true. Graph both sides. If the graphs are different, the equation is not an identity.
For example, is ${(\sin x+\cos x)^2=1}$ an identity?
It is not. To show that, one value of ${x}$ is enough. At ${x=\dfrac{\pi}{4}}$, the left side is $\left(\dfrac{\sqrt{2}}{2}+\dfrac{\sqrt{2}}{2}\right)^2=(\sqrt{2})^2=2$, not ${1}$.
But if the graphs look the same, that does not prove the identity. Graphs show only part of the picture and can hide small differences. A proof is needed.
How to prove an identity
To prove an identity, start with one side, and change it step by step, using known identities and algebra, until it becomes the other side.
- Start with the more complicated side. It has more to work with.
- Write it in terms of sine and cosine, if that helps.
- Use algebra: combine fractions, factor, expand, or multiply by a conjugate. (The conjugate of ${1-\sin x}$ is ${1+\sin x}$.)
- Keep the goal in sight. At each step, look at the other side and ask what would bring you closer to it.
Watch out: Do not treat an identity like an equation to solve. You may not add, multiply, or square both sides, because that assumes the identity is already true. Work on one side only.
Example 3: Prove the identity
$\cos\theta\,(\sec\theta-\cos\theta)=\sin^2\theta$
Solution:
Start with the left side. Write ${\sec\theta}$ as ${\dfrac{1}{\cos\theta}}$, and multiply out:
$\begin{align*}&\cos\theta\,(\sec\theta-\cos\theta)\\&=\cos\theta\cdot\dfrac{1}{\cos\theta}-\cos^2\theta\\&=1-\cos^2\theta\\&=\sin^2\theta\end{align*}$
This is the right side, so the identity is proved.
Example 4: Prove the identity
$\begin{align*}&\dfrac{1}{1-\sin x}+\dfrac{1}{1+\sin x}\\&=2\sec^2 x\end{align*}$
Solution:
Start with the left side, and combine the fractions. The common denominator is
$\begin{align*}&(1-\sin x)(1+\sin x)\\&=1-\sin^2 x\end{align*}$
$\begin{align*}&\dfrac{1}{1-\sin x}+\dfrac{1}{1+\sin x}\\&=\dfrac{(1+\sin x)+(1-\sin x)}{1-\sin^2 x}\\&=\dfrac{2}{\cos^2 x}\\&=2\sec^2 x\end{align*}$
Example 5: Prove the identity
$\dfrac{\cos x}{1-\sin x}=\dfrac{1+\sin x}{\cos x}$
Solution:
Start with the left side. Multiply the top and bottom by the conjugate ${1+\sin x}$. This turns the bottom into ${1-\sin^2 x}$, which is ${\cos^2 x}$:
$\begin{align*}&\dfrac{\cos x}{1-\sin x}\cdot\dfrac{1+\sin x}{1+\sin x}\\&=\dfrac{\cos x\,(1+\sin x)}{1-\sin^2 x}\\&=\dfrac{\cos x\,(1+\sin x)}{\cos^2 x}\\&=\dfrac{1+\sin x}{\cos x}\end{align*}$
In the last step, one ${\cos x}$ cancels from the top and the bottom.
Example 6: Prove the identity
${\tan^2 x-\sin^2 x=\tan^2 x\,\sin^2 x}$
Solution:
Start with the left side. Write ${\tan^2 x}$ as ${\dfrac{\sin^2 x}{\cos^2 x}}$ and combine:
$\begin{align*}&\dfrac{\sin^2 x}{\cos^2 x}-\sin^2 x\\&=\dfrac{\sin^2 x-\sin^2 x\cos^2 x}{\cos^2 x}\\&=\dfrac{\sin^2 x\,(1-\cos^2 x)}{\cos^2 x}\\&=\dfrac{\sin^2 x\cdot\sin^2 x}{\cos^2 x}\end{align*}$
Now split off one fraction:
$\dfrac{\sin^2 x}{\cos^2 x}\cdot\sin^2 x=\tan^2 x\,\sin^2 x$
Trigonometric substitution
Identities can also turn an algebra expression into a simpler trigonometric one. This trick is used a lot in calculus.
Example 7: Substitute ${x=\sin\theta}$, with ${-\dfrac{\pi}{2}\le\theta\le\dfrac{\pi}{2}}$, into ${\sqrt{1-x^2}}$, and simplify.
Solution:
$\begin{align*}\sqrt{1-x^2}&=\sqrt{1-\sin^2\theta}\\&=\sqrt{\cos^2\theta}=|\cos\theta|\end{align*}$
For ${\theta}$ between ${-\dfrac{\pi}{2}}$ and ${\dfrac{\pi}{2}}$, ${\cos\theta\ge 0}$. So ${|\cos\theta|=\cos\theta}$, and
${\sqrt{1-x^2}=\cos\theta}$
A right triangle shows the same thing when ${\theta}$ is acute:
Summary
- An identity is true for every value of the variable where both sides are defined.
- Know the reciprocal, quotient, Pythagorean, even-odd, and cofunction identities.
- To prove an identity, change one side step by step until it becomes the other side. Do not do the same thing to both sides.
- Useful moves: write in sines and cosines, combine fractions, factor, and multiply by a conjugate.
- To show an equation is not an identity, find one value of ${x}$ where it fails.