Which trigonometric identity is applicable to show $\sin\left(\frac{(2n+1)\pi}{2a}(x + a)\right) = \cos\left(\frac{(2n+1)\pi x}{2a} + n\pi \right)$? I'm confused with regards to the following and want to know which trigonometric identity is applicable here:
$$\sin\left(\frac{(2n+1)\pi}{2a}(x + a)\right) = \cos\left(\frac{(2n+1)\pi x}{2a} + n\pi \right)$$
where $a$ is a constant and $n \in \mathbb{Z}$.
Furthermore, I want to know why the above expression is equal to $$(-1)^n\cos\bigg(\frac{(2n+1)\pi x} {2a}\bigg).$$
Your first equation implies that
$$\sin\left(\phi+\frac{\pi}{2}\right)=\cos(\phi)$$
for any angle $\phi$, which is a defining identity of the trigonometric functions $\sin$ and $\cos$. Furthermore, you can easily see that both hand sides of your last equation are equal to
$$(-1)^n\cos(\phi)\,,$$
since $\cos(a+b)=\cos(a) \cos{b}-\sin(a) \sin{(b)}$.
Your first equation implies that
$$\sin\left(\phi+\frac{\pi}{2}\right)=\cos(\phi)$$
for any angle $\phi$, which is a defining identity of the trigonometric functions $\sin$ and $\cos$. Furthermore, you can easily see that both hand sides of your last equation are equal to
$$(-1)^n\cos(\phi)\,,$$
since $\cos(a+b)=\cos(a) \cos{b}-\sin(a) \sin{(b)}$.