Question

Sin(x+п/3)sin(x-п/3)=sinx.

Answer 1

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Answer 2

To prove the trigonometric identity $\sin(x + \frac{\pi}{3}) \sin(x - \frac{\pi}{3}) = \sin(x)$, we'll use the product-to-sum trigonometric identity. The product-to-sum identity states:

$\sin(A) \sin(B) = \frac{1}{2}[\cos(A-B) - \cos(A+B)]$

Let's use this identity to prove the given identity:

Starting with the left-hand side (LHS):

\begin{split}\sin(x + \frac{\pi}{3}) \sin(x - \frac{\pi}{3}) &= \frac{1}{2}[\cos(x - \frac{\pi}{3} - (x + \frac{\pi}{3})) - \cos(x - \frac{\pi}{3} + x + \frac{\pi}{3})] \\ &= \frac{1}{2}[\cos(x - x) - \cos(x + x)] \\ &= \frac{1}{2}[\cos(0) - \cos(2x)] \\ &= \frac{1}{2}[1 - \cos(2x)]\end{split}

Now, let's find the right-hand side (RHS):

$\sin(x) = \frac{1}{2}[\cos(x - (-x)) - \cos(x + (-x))] = \frac{1}{2}[1 - \cos(2x)]$

As we can see, both sides simplify to $\frac{1}{2}[1 - \cos(2x)]$, which are equal. Therefore, we have proved the given trigonometric identity: $\sin(x + \frac{\pi}{3}) \sin(x - \frac{\pi}{3}) = \sin(x)$.