# express

• Aug 24th 2009, 12:21 AM
thereddevils
express
If sin x + cos x =t , express $\sin^4x+\cos^4x$ in terms of t .

$\sin^4x=\frac{1}{4}(1-\cos2x)^2$

$=\frac{1}{4}(1-2\cos2x+\cos^22x)$

$\cos^4x=\frac{1}{4}(\cos2x+1)^2$

$
=\frac{1}{4}(1+2\cos2x+\cos^22x)
$

Putting them together .

$\frac{1}{4}(1-2\cos2x+\cos^22x+1+2\cos2x+\cos^22x)$

$=\frac{1}{2}(1-\cos^22x)$

I am not sure how to go on from here . . THanks for the hlep .
• Aug 24th 2009, 01:19 AM
red_dog
$\sin x+\cos x=t$

Square both members:

$\sin^2x+\cos^2x+2\sin x\cos x=t^2\Rightarrow 1+2\sin x\cos x=t^2\Rightarrow\sin x\cos x=\frac{t^2-1}{2}$

$\sin^4x+\cos^4x=(\sin^2x+\cos^2x)^2-2\sin^2x\cos^2x=$

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

Now replace $\sin x\cos x$ with $\frac{t^2-1}{2}$
• Aug 24th 2009, 03:11 AM
thereddevils
Quote:

Originally Posted by red_dog
$\sin x+\cos x=t$

Square both members:

$\sin^2x+\cos^2x+2\sin x\cos x=t^2\Rightarrow 1+2\sin x\cos x=t^2\Rightarrow\sin x\cos x=\frac{t^2-1}{2}$

$\sin^4x+\cos^4x=(\sin^2x+\cos^2x)^2-2\sin^2x\cos^2x=$

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

Now replace $\sin x\cos x$ with $\frac{t^2-1}{2}$

i complicated a simple problem .. thanks reddog .