Show that $\sin\left(\frac\pi3(x-2)\right)$ is equal to $\cos\left(\frac\pi3(x-7/2)\right)$ Show that 
$$\sin\left(\frac\pi3(x-2)\right)$$ 
is equal to 
$$\cos\left(\frac\pi3(x-7/2)\right)$$
I know that $\cos(x + \frac\pi2) = −\sin(x)$ but i'm not sure how i can apply it to this question.
 A: The equality $\sin \alpha=\cos\beta$ can be written
$$
\cos(\pi/2-\alpha)=\cos\beta
$$
which is satisfied when either
$$
\beta=\frac{\pi}{2}-\alpha+2k\pi \qquad(k \text{ integer})
$$
or
$$
\beta=\alpha-\frac{\pi}{2}+2k\pi \qquad(k \text{ integer})
$$
These can be rewritten respectively as
$$
\alpha+\beta=\frac{\pi}{2}+2k\pi \qquad(k \text{ integer})
$$
or
$$
\alpha-\beta=\frac{\pi}{2}+2k\pi \qquad(k \text{ integer})
$$
Now try with $\alpha=(\pi/3)(x-2)$ and $\beta=(\pi/3)(x-7/2)$; is one of the two equalities true?
A: Hint:
Try to use addition formulas. Let me get you started:
$$\sin\left(\frac\pi3(x-2)\right) = \sin\left(\frac{\pi x}{3} - \frac{2\pi}{3}\right) = \sin\frac{\pi x}{3}\cos\frac{2\pi}3 - \cos\frac{\pi x}3\sin\frac{2\pi}{3}=\\=-\frac12 \sin\frac{\pi x}{3} - \frac{\sqrt3}{2}\cos\frac{\pi x}{3}$$
Now, do a similar thing with the second expression.
A: The two angles $\frac{\pi}{3}(x-2)$ and $\frac{\pi}{3}(x-7/2)$ differs by $\pi/2$.
In fact,
$$ \frac{\pi}{3}(x-2) - \frac{\pi}{3}(x-\frac{7}{2}) = \frac{\pi}{3}x- \frac{2}{3}\pi - \frac{\pi}{3}x + \frac{7}{6}\pi = - \frac{7-4}{6}\pi = -\frac{\pi}{2}\, , $$
thus $\sin(\frac{\pi}{3}(x-2)) = \sin(\frac{\pi}{3}(x-7/2) - \frac{\pi}{2}) = \cos(\frac{\pi}{3}(x-7/2))$.
A: Maybe you can try
\begin{align*}\sin\left(\frac{\pi}{3}(x-2)\right) &= \sin\left(\frac{\pi}{3}x-\frac{2\pi}{3}\right)\\
&=\sin\left( \pi - \frac{\pi}{3}x+\frac{2\pi}{3}\right) \\
&=\sin\left(\frac{5\pi}{3}-\frac{\pi}{3}x\right) \\
&=\sin\left(\frac{\pi}{2}+\frac{7}{2}\frac{\pi}{3} -\frac{\pi}{3}x\right)\\
&=\cos\left(\frac{\pi}{3}\left(x-\frac{7}{2}\right)\right).
\end{align*}
