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I've been trying to prove the identity $$\sin2x + \sin2y = 2\sin(x + y)\cos(x - y).$$

So far I've used the identities based off of the compound angle formulas. I'm not quite sure if those identities would work with proving the above identity.

Thank you in advance.

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6 Answers

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\begin{align} 2\sin(x + y)\cos(x - y) &= 2(\sin x \cos y + \cos x \sin y)\cdot (\cos x \cos y + \sin x \sin y) \\ &= 2\sin x \cos x(\cos^2 y + \sin^2 y) + 2\sin y \cos y(\cos^2 x + \sin^2 x) \\ &= 2\sin x \cos x + 2\sin y \cos y \\ &= \sin 2x + \sin 2y \end{align}

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Hint:

Use the linearisation formula$$2\sin a\cos b=\sin(a+b)+\sin(a-b).$$

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Hint:

If $\sin(A+B),\sin(A-B)$ can be used

set $2x=A+B,2y=A-B\implies A= x+y, B=x-y$

and then expand $\sin(A\pm B)$

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Add the two compound angle formulas below,$$\sin(a+b) = \sin a \cos b + \cos a\sin b$$$$\sin(a-b) = \sin a \cos b - \cos a\sin b$$

to get,

$$\sin(a+b)+ \sin(a-b) = 2\sin a \cos b$$

Then, let $a=x+y$ and $b=x-y$ to obtain

$$\sin2x + \sin2y = 2\sin(x + y)\cos(x - y)$$

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Here is a proof of this identity using complex numbers.

Let $x, y \in \mathbb{R}^2$. By expanding

$$ e^{i(x+y)}e^{i(x-y)} = \big( \cos(x+y) + i\sin(x+y) \big) \big( \cos(x-y) + i\sin(x-y) \big),$$

you obtain:

$$ \Im\big( e^{i(x+y)}e^{i(x-y)} \big) = \cos(x+y)\sin(x-y) + \cos(x-y)\sin(x+y), $$

where $\Im(z)$ denotes the imaginary part of $z \in \mathbb{C}$. Therefore:

$$ \Im\big( e^{2ix} \big) = \sin(2x) = \cos(x+y)\sin(x-y) + \cos(x-y)\sin(x+y). \quad (\star) $$

By exchanging the role of $x$ and $y$ in $(\star)$, you also get:

$$ \Im\big( e^{2iy} \big) = \sin(2y) = -\cos(x+y) \sin(x-y) + \cos(x-y)\sin(x+y). \quad (\star\star)$$

Putting $(\star)$ and $(\star\star)$ together, you eventually get:

$$ \forall x, y \in \mathbb{R}^2, \; \sin(2x) + \sin(2y) = 2\cos(x-y)\sin(x+y). $$

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Hint:

Use factor formula, or also known as sum-to-product formula. Note that the formula works both ways i.e. can be interpreted as product-to-sum as well when read from right to left.

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