Justifying $\sin\left(\left(n+\frac{1}{2}\right)\pi\right) = (-1)^n$ without a calculator How do you justify this equation without a calculator?

$$\sin\left(\left(n+\frac{1}{2}\right)\pi\right) = (-1)^n$$
  where "n" is integers greater or equal to 1.

I tried using special triangles and identities, but I still couldn't figure it out.
 A: HINT:
If $n=2m,$  $$\sin\left(2m\pi+\dfrac\pi2\right)=\sin\dfrac\pi2=(-1)^{2m}$$
What if $n=2m+1$?

Alternatively use $\sin(A+B)$
A: By induction.
For $n=0$ one has $\sin{\pi\over 2}=1$
Assume that the identity is valid for $n-1$.
Let's write
$$\sin{\left(n+{1\over 2}\right)\pi}=\sin{\left(\pi+\left(n-1+{1\over 2}\right)\pi\right)}=\sin(\pi+\alpha)$$
with $\alpha=\left(n-1+{1\over 2}\right)\pi$. Now we know that $\forall \alpha,\,\sin(\pi+\alpha)=-\sin{\alpha}$ and therefore
$$\sin{\left(n+{1\over 2}\right)\pi}=-\sin{\left(n-1+{1\over 2}\right)\pi}=-(-1)^{n-1}=(-1)^n$$
A: Take the point $P=(1,0)$ in the "$x,y$" plane and rotate it counter-clockwise about the point $(0,0).$ When you have rotated $P$ thru an angle $A,$ the projection of $P$ onto the $y$-axis is  $(\sin A,0)$ because the arc-distance that $P$ has travelled along the circle is $A$. When $(A/\pi)-1/2$ is an integer, the point you have rotated to  is  $(1,0)$ if $n$ is even and is $(-1,0)$ if $n$ is odd ( because $\pi /2$ is one-fourth of the length of the circumference of a circle with radius $1$ .)
A: Draw the function in the cartesian plane. Look at where the points $(n+1/2)\pi$ lie. They just flip back and forth between the (1,0),(-1,0) points for different values of n.
