Trigonometry problem cosine identity

Question

Let $\cos^6\theta = a_6\cos6\theta+a_5\cos5\theta+a_4\cos4\theta+a_3\cos3\theta+a_2\cos2\theta+a_1\cos\theta+a_0$. Then $a_0$ is

(A) $0$ (B) $\frac{1}{32}$ (C) $\frac{15}{32}$ (D) $\frac{10}{32}$

Any hints on how to approach this?

Answer 1

To solve this question, you have a choice of using

(1) Top level – Through Fourier Analysis

(2) Advanced level – By complex number approach

(3) Intermediate level – Via compound angle formulas

(4) Elementary level (the most painful method) – Use brute force substitution

[Classifying into these levels is purely according to what grade the student is in.]

For (4), the logic is:-

Since the given ‘identity’ has 7 unknowns, you therefore need 7 independent equations. After solving most (if not all) these equations, you get the value of a_0 (hopefully, depending on whether the equations thus setup are independent or not).

These 7 equations (I do hope that they are independent) can be obtained by putting θ = 0, 30, 45, 60, 90, 120, 180 (or even 270) in succession.

Fortunately, some of the terms in these equations turn out to be zero (meaning that they are not that difficult to solve as seemed.) For example, when θ = 90, we have $0 = a_6(-1) + 0 + a_4(1) + 0 + a_2(-1) + 0 + a_0$

Enjoy.

Answer 2

Using \begin{align}&2\cos^2 x=\cos 2x +1, \\&4\cos^3x=\cos 3x+3\cos x, \\ \text{and}&2\cos a \cos b= \cos(a+b)+\cos(a-b),\end{align}

we obtain,

\begin{align}&\cos^6 x=(\cos^3 x)^2\\ \\=&\left(\dfrac{\cos 3x+3\cos x}{4}\right)^2\\ \\=&\dfrac1{16}(\cos^2 3x + 9\cos^2 x+6\cos x\cos 3x)\\ \\=&\dfrac{1}{16}\left(\dfrac{\cos 6x+1}{2}+9\dfrac{\cos 2x +1}2+3\cos 4x +3\cos 2x\right).\end{align}

It is seen that the constant term is $\dfrac{5}{16}$ or $\dfrac{10}{32}$.

(This of course, makes use of the question's assumption that the number $a_0$ is unique.)