# Cofunction Formulas with Problem Solution & Solved Example

## Cofunction Formulas

We often come across multiple functions in mathematics and one of the important methods is cofunction formulas in mathematics. Any function is called the cofunction if f(A) = g(B), or A and B are the complementary angles.

This is a special function that tells you relationship among inputs and outputs. Here, the input function is connected to output function somehow by means of a particular property and the trigonometric functions are one of them.

With cofunction formulas, you can quickly identify the relationship among various functions like sine, cosine, tangent, cotangent, secant and cosecant. Here, the value of a trigonometric function for an angle is equal to the value of cofunction of the complement.

### Co-function Identities

$\ \sin \theta =\cos\:(90-\theta)$
$\ \cos \theta =\sin\:(90-\theta)$
$\ \tan\theta =\cot\:(90-\theta)$
$\ \csc \theta =\sec\:(90-\theta)$
$\ \sec\theta =\csc\:(90-\theta)$
$\ \cot\theta =\tan\:(90-\theta)$

### Formula of Co-function

$\ \sin \left [\frac{\pi}{2}-\theta\right]=\cos\theta$
$\ \cos \left [\frac{\pi}{2}-\theta\right]=\sin \theta$
$\ \tan\left [\frac{\pi}{2}-\theta\right]=\cot \theta$
$\ \cot\left [\frac{\pi}{2}-\theta\right]=\tan\theta$
$\ \csc\left [\frac{\pi}{2}-\theta\right]=\sec\theta$
$\ \sec\left [\frac{\pi}{2}-\theta\right]=\csc\theta$

### Trigonometric Co-Functions

 $\ \sin\theta$ $\ \cos\theta$ $\ \tan\theta$ $\ \csc\theta$ $\ \sec\theta$ $\ \cot\theta$ $\ \sin\theta$ $\ \sin\theta$ $\ \pm \sqrt{1-\cos^{2}\theta}$ $\ \pm \frac{\tan\theta}{\sqrt{1+\tan ^{2}\theta}}$ $\ \frac{1}{\csc\theta}$ $\ \pm \frac{\sqrt{\sec^{2}-1}}{\sec\theta}$ $\ \pm \frac{1}{\sqrt{1+\cot^{2}\theta}}$ $\ \cos\theta$ $\ \pm \sqrt{1-\sin ^{2}\theta}$ $\ \cos\theta$ $\ \pm \frac{1}{\sqrt{1+\tan^{2}\theta}}$ $\ \pm \frac{\sqrt{\csc^{2}\theta-1}}{\csc\theta}$ $\ \frac{1}{\sec\theta}$ $\ \pm \frac{\cot\theta}{\sqrt{1+\cot^{2}\theta}}$ $\ \tan\theta$ $\ \pm \frac{\sin\theta}{\sqrt{1-\sin ^{2}}\theta}$ $\ \pm \frac{\sqrt{1-\cos^{2}\theta}}{\cos\theta}$ $\ \tan\theta$ $\ \pm \frac{1}{\sqrt{\csc^{2}\theta-1}}$ $\ \pm \sqrt{\sec^{2}\theta -1}$ $\ \frac{1}{\cot\theta}$ $\ \csc\theta$ $\ \frac{1}{\sin\theta}$ $\ \pm \frac{1}{\sqrt{1-\cos ^{2}\theta}}$ $\ \pm \frac{\sqrt{1+\tan ^{2}\theta}}{\tan\theta}$ $\ \csc\theta$ $\ \pm \frac{\sec\theta}{\sqrt{\sec^{2}\theta-1}}$ $\ \pm{\sqrt{1+\cot^{2}\theta}}$ $\ \sec\theta$ $\ \pm \frac{1}{\sqrt{1-\sin^{2}\theta}}$ $\ \frac{1}{\cos\theta}$ $\ \pm{\sqrt{1+\tan^{2}\theta}}$ $\ \pm \frac{\csc\theta}{\sqrt{\csc^{2}\theta-1}}$ $\ \sec\theta$ $\ \pm \frac{\sqrt{1+\cot^{2}\theta}}{\cot\theta}$ $\ \cot\theta$ $\ \pm \frac{\sqrt{1-\sin^{2}\theta}}{\sin\theta}$ $\ \pm \frac{\cos\theta}{\sqrt{1-\cos^{2}\theta}}$ $\ \frac{1}{\tan\theta}$ $\ \pm{\sqrt{\csc^{2}\theta-1}}$ $\ \pm \frac{1}{\sqrt{\sec^{2}\theta -1}}$ $\ \cot\theta$

If you will brush up the geometry concepts then a complement is defined as angles whose sum is always equal to the 90 degrees. Cofunctions in Trigonometry are function pairs like sine or cosine, tan or sec, Cot or cosecant etc. And the word “co” added in the beginning shows that there is something special about these functions.

Now, let us define the cofunctions in terms of θ. If one of angles is θ then the complementary angle would 90 – θ. This equation definitely makes sense because sum of three sides in a triangle is 180 and in case of right angle one angle must be equal to the 90 degrees. The concept may sound complicated in beginning but it could be made easier with little understanding and continuous practice.