Deduce that ;

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

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

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

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

$\displaystyle = 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) $