Problem

Problem

If cos⁡5θ=acos⁡θ+bcos⁡3θ+ccos⁡5θ+d\cos 5\theta = a \cos \theta + b \cos^3 \theta + c \cos^5 \theta + d , then a. a=20a = 20 b. b=−30b = -30 c. a+b+c=2a+b+c = 2 d. a+b+c+d=1a+b+c+d = 1
Intuitive explanation

This problem asks you to express cos⁡5θ\cos 5\theta as a polynomial in cos⁡θ\cos \theta—a classic application of Chebyshev polynomials. The key insight is to use the angle-addition formula or the recurrence relation cos⁡(nθ)\cos(n\theta) repeatedly: each time you add another θ\theta, you can rewrite the result using the identity cos⁡2θ=1−sin⁡2θ\cos^2 \theta = 1 - \sin^2 \theta to eliminate higher powers of sine. By systematically building up from cos⁡θ\cos \theta to cos⁡5θ\cos 5\theta, you'll discover that the coefficients a,b,c,da, b, c, d follow a predictable pattern tied to powers of cos⁡θ\cos \theta.

Intuitive explanation

This problem asks you to express cos⁡5θ\cos 5\theta as a polynomial in cos⁡θ\cos \theta—a classic application of Chebyshev polynomials. The key insight is to use the angle-addition formula or the recurrence relation cos⁡(nθ)\cos(n\theta) repeatedly: each time you add another θ\theta, you can rewrite the result using the identity cos⁡2θ=1−sin⁡2θ\cos^2 \theta = 1 - \sin^2 \theta to eliminate higher powers of sine. By systematically building up from cos⁡θ\cos \theta to cos⁡5θ\cos 5\theta, you'll discover that the coefficients a,b,c,da, b, c, d follow a predictable pattern tied to powers of cos⁡θ\cos \theta.

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