Show (sin⁡(θ)−cos⁡(2θ))2=1+sin⁡(θ)−sin⁡(3θ)−12cos⁡(2θ)+12cos⁡(4θ)

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kaluitagf · Accepted Answer

Method I. Write everything in terms of sin⁡(θ)  and  cos⁡(θ), and then compare. Tedious, but doable.  Method II. If you know what a Fourier series is, you may determine the coefficients on the right-hand side by integration.  Method III. First expand the left by (a+b)2=a2+2ab+b2  (sin⁡x−cos⁡(2x))2=sin2x+cos2(2x)−2sin⁡xcos⁡(2x)  Then use the double angle formulas:  sin2x=1−cos2x2,   cos2(2x)=1+cos⁡(4x)2  and the general product to sum formula:   sin⁡xcos⁡(2x)=12−sin⁡(x)+sin⁡(3x)

Louis Page · Accepted Answer

As we know the RHS, let&#039;s try to express cos2(2θ) in terms of cos⁡(4θ) (sin⁡θ−cos⁡2θ)2=sin2θ−2sin⁡θcos⁡2θ+cos22θ =sin2θ−2sin⁡θcos⁡2θ+1+cos⁡4θ2 Now cos⁡2θ=1−2sin2θwe use this together 4sin3θ (sin⁡θ−cos⁡2θ)2=sin2θ−2sin⁡θ+4sin3θ+12+cos⁡4θ2 =sin2θ+sin⁡θ−(3sin⁡θ−4sin3θ)+12+cos⁡4θ2 =sin⁡θ−sin⁡3θ+cos⁡4θ2+1−12+sin2θ =sin⁡θ−sin⁡3θ+cos⁡4θ2+1−1−2sin2θ2 =1+sin⁡θ−sin⁡3θ+cos⁡4θ2−cos⁡2θ2

Vasquez · Accepted Answer

Apply the identities cos⁡4θ=2cos2⁡2θ−1,    cos⁡2θ=1−2sin2⁡θ sin⁡3θ=sin⁡θ(2cos⁡2θ+1) in simplifying the RHS RHS=1+sin⁡θ−sin⁡θ(2cos⁡2θ+1)−12(1−2sin2⁡θ)+12(2cos2⁡2θ−1) =sin2⁡θ−2sin⁡θcos⁡2θ+cos2⁡2θ =(sin⁡θ−cos⁡2θ)2=LHS

Show (\sin(\theta)−\cos(2\theta))^2=1+\sin(\theta)−\sin(3\theta)−\frac12\cos(2\theta)+\frac12\cos(4\theta)

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