# 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.

• The statement is obvioulsy not true. The first expression has an $x$, and the second does not.
– 5xum
Commented Apr 28, 2015 at 9:06
• Oops sorry. I added the x now. Commented Apr 28, 2015 at 9:15
• You know that $\cos(y + \pi/2) = - \sin{y}$. Now choose $y = (\pi/3) (x - 2)$. Commented Apr 28, 2015 at 10:31

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?

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 $\sin(\frac{\pi}{3}(x-2)) = \sin(\frac{\pi}{3}(x-7/2) - \frac{\pi}{2}) = \cos(\frac{\pi}{3}(x-7/2))$.

• Second line evaluates to $\frac{\pi}2$. Commented Dec 20, 2021 at 12:23

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*}

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.