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Hello thereddevils

Quote:

Originally Posted by thereddevils

Show that $sin(\frac{\pi}{4}-\theta)=cos(\frac{\pi}{4}+\theta)$ and

$sin(\frac{\pi}{4}+\theta)=cos(\frac{\pi}{4}-\theta)$

i can do this part ..

Show that $sin^4(\frac{\pi}{4}-\theta)+sin^4(\frac{\pi}{4}+\theta)=\frac{3}{4}-\frac{1}{4}cos4\theta$

i cant do this part .

$\sin^4A = (\sin^2A)^2$

$=\Big(\tfrac12(1-\cos2A)\Big)^2$

$= \tfrac14(1-2\cos2A + \cos^22A)$

$=\tfrac14(1 - 2\cos2A +\tfrac12(\cos4A +1))$

$=\tfrac14(\tfrac32 - 2\cos2A +\tfrac12\cos4A)$

If we now replace $A$ by $(\tfrac{\pi}{4}+\theta)$ and then by $(\tfrac{\pi}{4}-\theta)$ , and add, we get:

$\sin^4(\tfrac{\pi}{4}+\theta)+\sin^4 (\tfrac{\pi}{4}-\theta)=\tfrac14\Big(3 - 2[\cos(\tfrac{\pi}{2}+2\theta) + \cos(\tfrac{\pi}{2}-2\theta)]+\tfrac12[\cos(\pi+4\theta)+\cos(\pi-4\theta)]\Big)$

Now use the formula for the sum of two cosines: $\cos A + \cos B = 2\cos\tfrac12(A+B)\cos\tfrac12(A-B)$

$\sin^4(\tfrac{\pi}{4}+\theta)+\sin^4 (\tfrac{\pi}{4}-\theta)=\tfrac14(3 - 4\cos\tfrac{\pi}{2}\cos2\theta+\cos\pi\cos4\theta)$

$= \tfrac34 -\tfrac14\cos4\theta$

Grandad