Trigonometic Identities

Printable View

November 16th 2006, 06:47 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
November 16th 2006, 06:53 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,
$\sin^2 x+\cos^2 x=1$
November 16th 2006, 06:57 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
November 16th 2006, 07:05 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,
$4\tan^2 x-1$

All times are GMT -8. The time now is 12:18 PM.

Copyright © 2005-2013 Math Help Forum. All rights reserved.

Copyright © 2005-2013 Math Help Forum. All rights reserved.