how do i prove (sin^4 x)-(cos^4 x)= (2sin^2x)-1
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Originally Posted by somanyquestions how do i prove (sin^4 x)-(cos^4 x)= (2sin^2x)-1 $\displaystyle \sin^4 x-\cos^4 x$ $\displaystyle (\sin^2 x)^2-(\cos^2 x)^2$ $\displaystyle (\sin^2 x)^2-(1-\sin^2 x)^2$ Expand....
Originally Posted by pickslides $\displaystyle \sin^4 x-\cos^4 x$ $\displaystyle (\sin^2 x)^2-(\cos^2 x)^2$ $\displaystyle (\sin^2 x)^2-(1-\sin^2 x)^2$ Expand.... ?? expand?
$\displaystyle \left [\sin^2 (x)\right ]^2-\left [ 1-\sin^2 (x)\right ]^2$ $\displaystyle =\left [sin^2(x)+(1-sin^2(x))\right ]\left [sin^2(x)-(1-sin^2(x))\right]$
Originally Posted by Stroodle $\displaystyle \left [\sin^2 (x)\right ]^2-\left [ 1-\sin^2 (x)\right ]^2$ $\displaystyle =\left [sin^2(x)+(1-sin^2(x))\right ]\left [sin^2(x)-(1-sin^2(x))\right]$ i have no idea what that means but this is what i did: sin4 x -(1- 2sin^2 x + sin4 x) sin4 x - 1+ 2sin^2 x - sin4 x sin4 x's cancel out -1+2sin^2 x = 2sin^2 x-1
Originally Posted by somanyquestions ?? expand? $\displaystyle (\sin^2 x)^2-(1-\sin^2 x)^2$ $\displaystyle (\sin^2 x)^2-(1-\sin^2 x)(1-\sin^2 x)$ $\displaystyle (\sin^2 x)^2-(1-2\sin^2 x+(\sin^2 x)^2)$ $\displaystyle (\sin^2 x)^2-1+2\sin^2 x-(\sin^2 x)^2$ finish...
That works too
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