what value this equal $\sin\left((2n-1)\frac{\pi}2\right)=$?

Question

The answer is $(-1)^n$

And why is this wrong?

$$\begin{align}\sin\left[(2n-1)\frac\pi2\right]&= sin(nπ-π/2)\\ &= \sin(n\pi) - \sin \left(\frac\pi2\right)\\ &= 0 - 1 = -1 \end{align}$$

Answer 1

If $\displaystyle\sin\left(A-B\right) = \sin A-\sin B$

$$\iff2\sin\frac{A-B}2\cos\frac{A-B}2 = 2\sin\frac{A-B}2\cos\frac{A+B}2$$

$$\iff2\sin\frac{A-B}2\left(\cos\frac{A-B}2-\cos\frac{A+B}2\right)=0$$

$$\iff2\sin\frac{A-B}2\left(2\sin\frac A2\sin\frac B2\right)=0$$

Now, if $\sin x=0,x=m\pi$ where $m$ is any integer

Clearly, the supposed identity does not hold true for all $A,B$

In fact, $$\sin\left(A-B\right)=\sin A\cos B-\cos A\sin B$$

So, $$\sin\left(n\pi-\dfrac\pi2\right)=\sin(n\pi)\cos\dfrac\pi2-\cos(n\pi)\sin\dfrac\pi2=-\cos(n\pi)=-(-1)^n$$ for integer $n$

Answer 2

$$\sin\left(n\pi-\frac{\pi}{2}\right)=-\cos(n\pi)=(-1)^{n+1}$$