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Question: If $\sin 2\theta + \sin 2\phi = \frac{1}{2}$ and $\cos 2\theta + \cos 2\phi = \frac{3}{2}$, $\cos^2...

If sin⁡2θ+sin⁡2ϕ=12\sin 2\theta + \sin 2\phi = \frac{1}{2} and cos⁡2θ+cos⁡2ϕ=32\cos 2\theta + \cos 2\phi = \frac{3}{2},

cos⁡2(θ−ϕ)=\cos^2(\theta - \phi) =

A

38\frac{3}{8}

B

58\frac{5}{8}

C

34\frac{3}{4}

D

54\frac{5}{4}

Answer

58\frac{5}{8}

Explanation

Solution

We are given:

  1. sin⁡2θ+sin⁡2ϕ=12\sin 2\theta + \sin 2\phi = \frac{1}{2}
  2. cos⁡2θ+cos⁡2ϕ=32\cos 2\theta + \cos 2\phi = \frac{3}{2}

Using sum-to-product identities:

sin⁡A+sin⁡B=2sin⁡(A+B2)cos⁡(A−B2)\sin A + \sin B = 2 \sin \left(\frac{A+B}{2}\right) \cos \left(\frac{A-B}{2}\right)

cos⁡A+cos⁡B=2cos⁡(A+B2)cos⁡(A−B2)\cos A + \cos B = 2 \cos \left(\frac{A+B}{2}\right) \cos \left(\frac{A-B}{2}\right)

Applying these identities:

2sin⁡(θ+ϕ)cos⁡(θ−ϕ)=122 \sin (\theta+\phi) \cos (\theta-\phi) = \frac{1}{2} (Equation 3)

2cos⁡(θ+ϕ)cos⁡(θ−ϕ)=322 \cos (\theta+\phi) \cos (\theta-\phi) = \frac{3}{2} (Equation 4)

Squaring both equations:

4sin⁡2(θ+ϕ)cos⁡2(θ−ϕ)=144 \sin^2 (\theta+\phi) \cos^2 (\theta-\phi) = \frac{1}{4} (Equation 5)

4cos⁡2(θ+ϕ)cos⁡2(θ−ϕ)=944 \cos^2 (\theta+\phi) \cos^2 (\theta-\phi) = \frac{9}{4} (Equation 6)

Adding Equation 5 and Equation 6:

4sin⁡2(θ+ϕ)cos⁡2(θ−ϕ)+4cos⁡2(θ+ϕ)cos⁡2(θ−ϕ)=14+944 \sin^2 (\theta+\phi) \cos^2 (\theta-\phi) + 4 \cos^2 (\theta+\phi) \cos^2 (\theta-\phi) = \frac{1}{4} + \frac{9}{4}

Factoring:

4cos⁡2(θ−ϕ)[sin⁡2(θ+ϕ)+cos⁡2(θ+ϕ)]=1044 \cos^2 (\theta-\phi) [\sin^2 (\theta+\phi) + \cos^2 (\theta+\phi)] = \frac{10}{4}

Using the identity sin⁡2x+cos⁡2x=1\sin^2 x + \cos^2 x = 1:

4cos⁡2(θ−ϕ)=1044 \cos^2 (\theta-\phi) = \frac{10}{4}

Solving for cos⁡2(θ−ϕ)\cos^2 (\theta-\phi):

cos⁡2(θ−ϕ)=1016=58\cos^2 (\theta-\phi) = \frac{10}{16} = \frac{5}{8}