# Thread: trig indentity proof help,

1. ## trig indentity proof help,

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

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

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

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

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

2. I would try

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

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

Therefore

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

Now expand and group like terms.

3. Actually you were just one step away from the answer.

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

4. 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 :

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

5. 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