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  1. #1
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    trigo show question

    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 .
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  2. #2
    MHF Contributor
    Grandad's Avatar
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    Trig proof

    Hello thereddevils
    Quote Originally Posted by thereddevils View Post
    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
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