This post is part of a series of posts on Trigonometry. To see all the posts, click on the tag #TrigonometryTutorials below. This is the post you should read before you read this.

Here are a few trig identities that are very important to remember:

**Identity 1**:
\[\sin^2 x+ \cos^2 x = 1\]

Proof: Use the Definition of the trig functions: \(\sin x = \frac oh\) and \(\cos \frac ah\)

\[\left(\frac oh\right)^2+\left(\frac ah\right)^2= 1\] \[\frac {o^2}{h^2} + \frac {a^2}{h^2} = 1\]

Multiply both sides by \(h^2\)

\[a^2+o^2=h^2\]

This is just the Pythagorean Theorem! So this holds true. \(\blacksquare\)

A variation of Identity 1 can be achieved by:

dividing both sides by \(\sin^2 x\)

\[1+ \cot^2 x = \csc x\]

dividing both sides by \(\cos^2 x\)

\[ \tan^2 x +1 = \sec x\]

**Identity 2**: \(\sin x = \cos (90^\circ-x)\)

In a Right angled triangle, if one of the angles is \(x\), The other angle will be \(90-x\) (the angle other than the right angle, that is.). The opposite angle of one angle is the adjacent of the other angle, therefore \(\sin x = \cos (90^\circ-x)\).

Using the same reasoning, these other identities can be obtained:

\[\cos x = \sin (90-x)\] \[\sec x = \csc (90-x)\] \[\csc x = \sec (90-x)\] \[\tan x = \cot(90-x)\] \[\cot x = \tan (90-x)\]

The Next post in this series is here

## Comments

There are no comments in this discussion.