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wazhma1991 · Mar 24, 2009, 05:23 AM

Hi galactus how shoud I solve this :confused:

sin^2 20=cos^2 20-cos^2 40

galactus · Mar 24, 2009, 08:47 AM

Are you solving it or proving they are the same?

wazhma1991 · Mar 25, 2009, 05:57 AM

Hi I want to solved I know that is right but how?

galactus · Mar 25, 2009, 07:07 AM

$sin^{2}(2x)=cos^{2}(2x)-cos^{2}(4x)$

That is not an identity. They are not equivalent.

That's what leads me to believe you were asked to solve for theta perhaps.

But, to solve the equation, notice the identity $sin(2x)=2sin(x)cos(x)$

Therefore, $sin^{2}(2x)=4sin^{2}(x)cos^{2}(x)$

Another identity: $cos(2x)=cos^{2}(x)-sin^{2}(x)=1-2sin^{2}(x)=2cos^{2}(x)-1$

$cos(4x)=cos^{2}(2x)-sin^{2}(2x)=1-2sin^{2}(2x)=2cos^{2}(2x)-1$

Try $x=\pm\frac{\pi}{2}, \;\ \pm{\pi}, \;\ \pm\frac{3\pi}{8}$

Play around with various identities until you can whittle it down enough to solve for x.