Question

Cos 6 theta in terms of cos theta

Answer 1

Answer:

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Step-by-step explanation:

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Answer 2

Answer:

we have cos6θ=cos(2×3θ)=2cos23θ−1=2(4cos3θ−3cosθ)2−1=32cos6θ−48cos4θ+18cos2θ−1. ADDED: From the definition of the Chebyshev polynomials Tn(x)=cos(narccosx)⇔Tn(cosθ)=cosnθ,θ=arccosx, we get T1(x)=cos(arccosx)=xT2(x)=cos(2arccosx)