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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.

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

4 Answers 4

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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?

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

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

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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.

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