Why when we add or substract $\alpha$ to a odd multiple of $\frac{\pi}{2}$ inside $\sin/\cos$ we get cofunction?

Question

Why when we add or substract $\alpha$ to a odd multiple of $\frac{\pi}{2}$ inside $\sin/\cos$ we get cofunction and when we do same thing with multiple of $\pi$ we get just same function but with sign depending on quarter? Where the transformation for $\frac{\pi}{2}$ comes from?

Answer 1

Recall the sum/difference formulas:

$$\sin(x \pm y) = \sin x \cos y \pm \cos x \sin y$$ $$\cos(x \pm y) = \cos x \cos y \mp \sin x \sin y$$

Now, let $y = \frac{\pi}{2}n$ for some odd integer $n$.

$$\sin(x + \frac{\pi}{2}n) = \sin x \cos (\frac{\pi}{2}n) + \cos x \sin (\frac{\pi}{2}n)$$ $$\cos(x + \frac{\pi}{2}n) = \cos x \cos (\frac{\pi}{2}n) - \sin x \sin (\frac{\pi}{2}n)$$

But when $n$ is odd, $\cos(\frac{\pi}{2}n) = 0$, so this simplifies to:

$$\sin(x + \frac{\pi}{2}n) = \cos x \sin (\frac{\pi}{2}n)$$ $$\cos(x + \frac{\pi}{2}n) = - \sin x \sin (\frac{\pi}{2}n)$$

Since $n$ is odd, there is an integer $k$ such that $n = 2k + 1$. So:

$$\sin (\frac{\pi}{2}n) = \sin (\frac{\pi}{2}(2k + 1))$$ $$= \sin(\pi k + \frac{\pi}{2})$$ $$= \sin(\pi k)\cos(\frac{\pi}{2}) + \cos(\pi k)\sin(\frac{\pi}{2})$$ $$= 0 \times 0 + (-1)^k \times 1$$ $$= (-1)^k$$

Plugging into our previous set of equations gives:

$$\sin(x + \frac{\pi}{2}(2k + 1)) = (-1)^k\cos x$$ $$\cos(x + \frac{\pi}{2}(2k + 1)) = (-1)^{k+1}\sin x$$

Answer 2

Even if OP does not know matrix representation of a rotation, it may be useful for some other readers.

Let $r(x)$ and $r(y)$ two rotations. In a direct orthonormal basis of the plane,$$r(x)=\begin{bmatrix}\cos x & -\sin x \\ \sin x& \cos x\end{bmatrix}, r(y)=\begin{bmatrix}\cos y & -\sin y \\ \sin y& \cos y\end{bmatrix}$$So, $$\begin{bmatrix}\cos x+y & -\sin x+y \\ \sin x+y& \cos x+y\end{bmatrix}=r(x+y):=r(x)r(y)$$$$r(x-y):=r(x)r(y)^{-1}$$And we find the formulas provided by @Dan.