Trigonometry: Basic Trig Identities

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: sin2x+cos2x=1\sin^2 x+ \cos^2 x = 1

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

(oh)2+(ah)2=1\left(\frac oh\right)^2+\left(\frac ah\right)^2= 1 o2h2+a2h2=1\frac {o^2}{h^2} + \frac {a^2}{h^2} = 1

Multiply both sides by h2h^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 sin2x\sin^2 x

1+cot2x=cscx1+ \cot^2 x = \csc x

dividing both sides by cos2x\cos^2 x

tan2x+1=secx \tan^2 x +1 = \sec x

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

In a Right angled triangle, if one of the angles is xx, The other angle will be 90x90-x (the angle other than the right angle, that is.). The opposite angle of one angle is the adjacent of the other angle, therefore sinx=cos(90x)\sin x = \cos (90^\circ-x).

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

cosx=sin(90x)\cos x = \sin (90-x) secx=csc(90x)\sec x = \csc (90-x) cscx=sec(90x)\csc x = \sec (90-x) tanx=cot(90x)\tan x = \cot(90-x) cotx=tan(90x)\cot x = \tan (90-x)

The Next post in this series is here

Note by Yan Yau Cheng
7 years, 6 months ago

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