Problem 2

Let

a_1,a_2,\ldots,a_n

be real constants and let

x

be real. Define

f(x)=\cos(a_1+x)+\frac12\cos(a_2+x)+\cdots+\frac1{2^{n-1}}\cos(a_n+x)

. Given that

f(x_1)=f(x_2)=0

, prove that

x_2-x_1=m\pi

for some integer

m

Step 4 of 4: Compare the zeros

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In plain words

Successive zeros of a cosine are separated by half-turns, exactly integer multiples of $\pi$ .

\cos(x_i+\delta)=0\Rightarrow x_i+\delta=\frac\pi2+k_i\pi\quad(i=1,2),\qquad x_2-x_1=(k_2-k_1)\pi

The unit circle marks a zero of cosine at a quarter-turn.

Detailed analysis

Since $R>0$ , the hypotheses imply $\cos(x_i+\delta)=0$ for $i=1,2$ . The zeros of cosine are $\frac\pi2+k\pi$ , so subtracting the two equations gives $x_2-x_1=(k_2-k_1)\pi$ , as required.