trig indentity proof help,

**Tweety** · December 15th 2009, 11:54 AM

Deduce that ;

$sin^{4}A + cos^{4}A = \frac{1}{4} (3 + cos4A)$

Can someone start me off please?

Thanks!

this is part 'b' of the question, I just realised you might actually need the first part, which I have already done.

part a is;

use the identity $sin^{2}A + cos^{2}A = 1$ to that show that $sin^{4}A + cos^{4}A = \frac{1}{2}(2-sin^{2}2A)$

$(sin^{2}A + cos^{2}A )^{2}= 1$

$sin^{4} + cos^{4} + 2sin^{2}A cos^{2}A = 1$

$sin^4 + cos^{4} = 1 - 2sin^{2}Acos^{2}A$

$= 1-\frac{1}{2} ( 4sin^2Acos^{2}A) = 1-\frac{1}{2} ( 2sinAcosA)^{2} = 1-\frac{1}{2}sin^{2}2A = \frac{1}{2} ( 2-sin^{2}2A)$

**pickslides** · December 15th 2009, 12:09 PM

I would try

$\sin^2(x) = \frac{1-\cos(2x)}{2}$

$\cos^2(x) = \frac{1+\cos(2x)}{2}$

Therefore

$<br /> \sin^4(x)+\cos^4(x) = \left(\frac{1-\cos(2x)}{2}\right)^2+\left(\frac{1+\cos(2x)}{2}\r ight)^2<br />$

Now expand and group like terms.

**Isomorphism** · December 15th 2009, 12:22 PM

Actually you were just one step away from the answer.

Use this:
$2(2 - \sin^2 2A) = 3 + (1 - 2\sin^2 2A) = 3 + \cos 4A \implies 2 - \sin^2 2A = \frac12(3 + \cos 4A)$

**ldx2** · December 15th 2009, 12:36 PM

In your part A question you didn't show that :

$sin^{4}A + cos^{4}A = \frac{1}{2}(2-sin^{2}A)$

You got this solution :

$\frac{1}{2} ( 2-sin^{2}2A)$

**Tweety** · December 15th 2009, 12:58 PM

Originally Posted by ldx2

In your part A question you didn't show that :

$sin^{4}A + cos^{4}A = \frac{1}{2}(2-sin^{2}A)$

You got this solution :

$\frac{1}{2} ( 2-sin^{2}2A)$

oh sorry, its meant to be 2A, I mis-typed

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