1
$\begingroup$

Prove $\sin{2x}+\sin{4x}+\sin{6x}=4\cos{x}\cos{2x}\sin{3x}$

I have reached the point where the LHS equation has turned into $2\cos{x}\cos2x\sin{x}(2\sin2x+1)$

But I have no idea how to turn $\sin{x}(2\sin2x+1)$ into $2\sin3x$

A quicker method if it exists would be greatly appreciated

Thanks in advance

$\endgroup$
2
$\begingroup$

To solve this problem, we can use the identities:

$$ \sin A + \sin B = 2\sin \frac{A+B}{2} \cos \frac{A - B}{2}, $$

$$ \cos A + \cos B = 2\cos \frac{A+B}{2} \cos \frac{A - B}{2}, $$

and

$$ \sin 2\phi = 2\sin \phi \cos \phi. $$

Going back to the question,

$ \text{LHS} = \sin 2x + \sin 4x + \sin 6x \\ = 2\sin 3x \cos x + \sin 6x \\ = 2\sin 3x \cos x + 2\sin 3x \cos 3x \\ = 2\sin 3x (\cos x + \cos 3x) \\ = 2\sin 3x \times 2\cos 2x \cos x \\ = 4\cos x \cos 2x \sin 3x \\ = \text{RHS}. $

Hence, proved.

$\endgroup$
1
$\begingroup$

Let me try. $$\sin 2x + \sin 6x + \sin 4x = 2\sin 4x \cos 2x + 2\sin 2x\cos 2x = 2\cos 2x (\sin 4x + \sin 2x) = 4\cos 2x \sin 3x \cos x$$

$\endgroup$
1
1
$\begingroup$

$$ \sin 2x + \sin 4x = \sin(3x-x) +\sin(3x+x) = 2 \sin 3x \cos x \\ $$ and $$ \sin 6x = 2 \sin 3x \cos 3x $$ so $$ \sin 2x + \sin 4x +\sin 6x = 2\sin 3x (\cos x + \cos 3x) $$ but $$ \cos x + \cos 3x = \cos(2x-x) + \cos (2x+x) = 2\cos x \cos 2x $$

$\endgroup$
1
$\begingroup$

From Euler's identity, $\sin(nx) = \frac{1}{2i}(e^{inx} - e^{-inx})$ and $\cos(nx) = \frac{1}{2}(e^{inx} + e^{-inx})$ Therefore:

$$4\cos(x)\cos(2x)\sin(3x) = \frac{1}{2i}(e^{ix}+e^{-ix})(e^{2ix}+e^{-2ix})(e^{3ix}-e^{-3ix})$$

$$=\frac{1}{2i}(e^{3ix} + e^{-ix} + e^{ix} + e^{-3ix})(e^{3ix} - e^{-3ix})$$

$$=\frac{1}{2i}(e^{6ix} - e^{-6ix} + e^{4ix} - e^{-4ix} + e^{2ix} - e^{-2ix})$$

$$=\sin(2x)+\sin(4x)+\sin(6x)$$

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.