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Miscellaneous Examples · Example 22

Q.Prove that cos⁡2x+cos⁡2(x+π3)+cos⁡2(x−π3)=32\cos^2 x + \cos^2\left(x + \frac{\pi}{3}\right) + \cos^2\left(x - \frac{\pi}{3}\right) = \frac{3}{2}.

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The identity is proved by expanding each cosine square using the double-angle formula cos⁡2θ=1+cos⁡2θ2\cos^2\theta = \frac{1+\cos 2\theta}{2}, then summing the three terms. The cosine terms cancel out, leaving only the constant 32\frac{3}{2}.

The key insight here is that when you have a sum of squares of cosines at angles that are symmetrically spaced, the double-angle trick turns the problem into a sum of cosines of 2x2x, 2x+2π/32x + 2\pi/3, and 2x−2π/32x - 2\pi/3. Those three cosines sum to zero — a classic result from phasor addition or from the identity cos⁡A+cos⁡B=2cos⁡A+B2cos⁡A−B2\cos A + \cos B = 2\cos\frac{A+B}{2}\cos\frac{A-B}{2}.

Let’s work through it.

  1. Apply the double-angle identity For any angle θ\theta, cos⁡2θ=1+cos⁡2θ2\cos^2\theta = \frac{1 + \cos 2\theta}{2}. So the left-hand side becomes:

1+cos⁡2x2+1+cos⁡(2x+2π3)2+1+cos⁡(2x−2π3)2\frac{1 + \cos 2x}{2} + \frac{1 + \cos\left(2x + \frac{2\pi}{3}\right)}{2} + \frac{1 + \cos\left(2x - \frac{2\pi}{3}\right)}{2}

  1. Separate the constants and the cosines Each term contributes a 12\frac{1}{2}, so the constant part is 32\frac{3}{2}. The cosine part is:

12[cos⁡2x+cos⁡(2x+2π3)+cos⁡(2x−2π3)]\frac{1}{2}\left[ \cos 2x + \cos\left(2x + \frac{2\pi}{3}\right) + \cos\left(2x - \frac{2\pi}{3}\right) \right]

  1. Simplify the sum of cosines Use the sum-to-product formula: cos⁡A+cos⁡B=2cos⁡A+B2cos⁡A−B2\cos A + \cos B = 2\cos\frac{A+B}{2}\cos\frac{A-B}{2}. Pair the last two terms:

cos⁡(2x+2π3)+cos⁡(2x−2π3)=2cos⁡(2x)cos⁡(2π3)\cos\left(2x + \frac{2\pi}{3}\right) + \cos\left(2x - \frac{2\pi}{3}\right) = 2\cos(2x)\cos\left(\frac{2\pi}{3}\right)

Since cos⁡(2π3)=−12\cos\left(\frac{2\pi}{3}\right) = -\frac{1}{2}, this becomes:

2cos⁡(2x)⋅(−12)=−cos⁡(2x)2\cos(2x) \cdot \left(-\frac{1}{2}\right) = -\cos(2x)

  1. Add the remaining term Now the full sum of cosines is:

cos⁡2x+[−cos⁡2x]=0\cos 2x + \left[ -\cos 2x \right] = 0

So the entire cosine contribution vanishes. …

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