# Trigonometic Identities

• Nov 16th 2006, 07:47 PM
asiankatt
Trigonometic Identities
Hi there .. I'm not sure how to approach this question and I would greatly appreciate it if someone could help me :)

Thanks a bunch .

Here's the question :

Prove the identity .

sin^4x - cos^4x = sin^2x - cos^2x
• Nov 16th 2006, 07:53 PM
ThePerfectHacker
Quote:

Originally Posted by asiankatt
Hi there .. I'm not sure how to approach this question and I would greatly appreciate it if someone could help me :)

Thanks a bunch .

Here's the question :

Prove the identity .

sin^4x - cos^4x = sin^2x - cos^2x

This is very simple.
Factor the left hand side,
$x^4-y^4=(x^2-y^2)(x^2+y^2)$
Thus,
$(\sin^2 x+\cos^2 x)(\sin^2 x-\cos^2 x)$
But,
$\sin^2 x+\cos^2 x=1$
• Nov 16th 2006, 07:57 PM
asiankatt
=)
Thanks a lot =) Now that I look at it .. it is really simple :S

.... I have another question if you dont mind answering it ...

Prove the indentity .

4/cos^2x - 5 = 4tan^2 x - 1
• Nov 16th 2006, 08:05 PM
ThePerfectHacker
Quote:

Originally Posted by asiankatt
Thanks a lot =) Now that I look at it .. it is really simple :S

.... I have another question if you dont mind answering it ...

Prove the indentity .

4/cos^2x - 5 = 4tan^2 x - 1

Express left hand side as,
$\frac{4}{\cos^2 x}-4-1$
Factor, and use reciprocal identity,
$4(\sec^2 x-1)-1$
But,
$\tan^2 x+1=\sec^2 x$
Thus,
$4\tan^2 x-1$