# express

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

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

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

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

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

Putting them together .

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

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

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

Square both members:

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

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

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

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

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

Square both members:

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

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

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

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

i complicated a simple problem .. thanks reddog .
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