express

**thereddevils** · August 24th 2009, 12:21 AM

If sin x + cos x =t , express $\sin^4x+\cos^4x$ in terms of t .

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

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

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

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

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 .

**red_dog** · August 24th 2009, 01:19 AM

$\sin x+\cos x=t$

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 $\frac{t^2-1}{2}$

**thereddevils** · August 24th 2009, 03:11 AM

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 .

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