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Trigonometry

Hi galactus how shoud I solve this :confused:

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

Are you solving it or proving they are the same?

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

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

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.