What rule is used when calculating $\cos(n \cdot \pi)\cdot\cos(m \cdot \pi)$

Question

I can figure out how you would calculate the product of the cosines of $\pi$ when there are no coefficients to get $1$ (trivially). I cannot however understand what rule is used/how we arrive at $(-1)^{(n+m)}$ for when we have two different coefficients of $\pi$, namely $n$ and $m$.

Answer 1

We have that

• $\cos(n \cdot \pi)=1$ for $n$ even
• $\cos(n \cdot \pi)=-1$ for $n$ odd

that is

• $\cos(n \cdot \pi)=(-1)^n$

thus

$$\cos(n \cdot \pi)\cdot \cos(m \cdot \pi)=(-1)^n(-1)^m=(-1)^{n+m}$$

Answer 2

The cosine function has period $2\pi$: $\cos(x+2\pi)=\cos x$ for all $x$. Therefore $$1=\cos0=\cos2\pi=\cos4\pi=\cdots$$ and $$-1=\cos\pi=\cos3\pi=\cos5\pi=\cdots.$$ Then $\cos n\pi=1$ when $n$ is even, and $\cos n\pi=-1$ if $n$ is odd.

In short, $\cos n\pi=(-1)^n$ for integers $n$.

Answer 3

$\cos \pi = -1$ and $\cos 2 \pi = 1$. Which means, every odd multiple of $\pi$ gives $-1$ as the value for $\cos$ and every even multiple of $\pi$ gives the value $1$. Thus, in general, we write

$$\cos n \pi = \left( -1 \right)^n$$

which settles the problem of $n$ begin odd or even. Therefore, in your question, what we have is

$$\cos m \pi \cdot \cos n \pi = \left( -1 \right)^m \cdot \left( -1 \right)^n = \left( -1 \right)^{n + m}$$