1. ## Trigonometric equations

In my book following result is shown but don't know where it come from.
1.2cos^2(2kpie)=2 (where k is an integer)
2.cos(kpie)=(-1)^k
3.[sin(kpie)cos(theta)-cos(kpie)sin(theta)]=[(-1)^k-1 sin(theta)]
In all these cases I just don't know how the results came.Please explain it to me.

2. Hi roshanhero,

1) The cosine function is periodic with period $2\pi$ and thus $\cos (2k\pi)=\cos (2\pi)~\forall k \in \mathbb{Z}$ , hence $2\cos^2 (2k\pi)=2\cos^2 (2\pi)~\forall k \in \mathbb{Z}$ The conclusion follows.

2) If $k$ is even then $k=2n$ for some $n$ and hence $\cos (k\pi)=\cos (2n\pi)=\cos (2\pi)~\forall n \in \mathbb{Z}$ (which follows from above). Notice that $(-1)^k=[(-1)^2]^n=1^n=1~\forall n \in \mathbb{Z}$ Thus if $k$ is even then the statement is true since $\cos (2\pi)=1$ Note that due to the cosine function's periodic attributes $\cos (\psi+\pi)\equiv-\cos (\psi)$ . If $k$ is odd then $k=2r+1$ for some $r$ and hence $\cos (k\pi)=\cos [(2r+1)\pi]=\cos (2r\pi+\pi)=-\cos (2r\pi) = -\cos (2\pi)~~\forall r \in \mathbb{Z}$ Notice that $(-1)^k=(-1)^{2r+1}=(-1)\cdot(-1)^{2r}=-1$ Thus if $k$ is odd then the statement is true since $-\cos (2\pi)=-1$ hence the statement is true $\forall k \in \mathbb{Z}$

3) I have to go so I will leave this one until later unless of course someone else would like to fill in the blanks.

3. I'm back.

3) Note that $\sin (\theta \pm \phi) \equiv \sin (\theta) \cdot \cos (\phi) \pm \cos (\theta) \cdot \sin (\phi)$ Following from this we can see that $\sin (k\pi) \cdot \cos (\theta) - \cos (k\pi) \cdot sin (\theta) \equiv \sin (k\pi - \theta)$ Like previously we will split this into two seperate cases. If $k$ is even then $k=2n$ for some $n$ and hence $\sin (k\pi - \theta) = \sin (-\theta + 2n\pi)$ The sine function is also periodic with period $2\pi$ and thus $\sin(\phi + 2x\pi) \equiv \sin (\phi)~ \forall x \in \mathbb{Z}$ Also note that $\sin(-\psi) \equiv -\sin (\psi)$ Knowing these two bits of information we can then see that $\sin (-\theta + 2n\pi) \equiv \sin (-\theta) \equiv -\sin (\theta)$ Notice that $(-1)^{k-1} \cdot \sin (\theta) = \frac{(-1)^{2n}}{-1} \cdot \sin (\theta) = -\sin (\theta)~\forall n \in \mathbb{Z}$ Hence the statement is true if $k$ is even since rather trivially $-\sin (\theta) = -\sin (\theta)$ .

Now if $k$ is odd then $k=2r+1$ for some $r$ and thus $\sin (k\pi - \theta) = \sin [-\theta + (2n+1) \pi] = \sin (-\theta + 2n \pi + \pi) = \sin (-\theta + \pi)$ Note that the sine function has some of the same attributes as the cosine function in that $\sin (\psi +\pi) = -\sin (\psi)$ Hence using this results implies that $\sin (-\theta + \pi) = -\sin (-\theta) = \sin (\theta)$ Notice that $(-1)^{k-1} \cdot \sin (\theta) = (-1)^{2r} \cdot \sin (\theta) = \sin (\theta)~\forall r \in \mathbb {Z}$ Hence the statement is true if $k$ is odd since again rather trivially $\sin (\theta) \equiv \sin (\theta)$ and thus the statement is true $~\forall k \in \mathbb {Z}$

$Q.E.D.$

Thats how the results came.