How do you prove (sin2x + sin2y)/(cos2x + cos2y) = tan(x + y)?

1 Answer
Dec 17, 2015

With the sum identities

• $\sin \left(x\right) + \sin \left(y\right) = 2 \sin \left(\frac{x + y}{2}\right) \cos \left(\frac{x - y}{2}\right)$

• $\cos \left(x\right) + \cos \left(y\right) = 2 \cos \left(\frac{x + y}{2}\right) \cos \left(\frac{x - y}{2}\right)$

we have

$\frac{\sin \left(2 x\right) + \sin \left(2 y\right)}{\cos \left(2 x\right) + \cos \left(2 y\right)} = \frac{2 \sin \left(\frac{2 x + 2 y}{2}\right) \cos \left(\frac{2 x - 2 y}{2}\right)}{2 \cos \left(\frac{2 x + 2 y}{2}\right) \cos \left(\frac{2 x - 2 y}{2}\right)}$

$= \sin \frac{x + y}{\cos} \left(x + y\right)$

$= \tan \left(x + y\right)$

In the case that the sum identities above are not given, here is a short proof of them using the angle-addition formulas:

Let $a = \frac{x + y}{2}$ and $b = \frac{x - y}{2}$

Then

$\sin \left(x\right) + \sin \left(y\right) = \sin \left(a + b\right) + \sin \left(a - b\right)$

$= \sin \left(a\right) \cos \left(b\right) + \cos \left(a\right) \sin \left(b\right) + \sin \left(a\right) \cos \left(b\right) - \cos \left(a\right) \sin \left(b\right)$

$= 2 \sin \left(a\right) \cos \left(b\right)$

$= 2 \sin \left(\frac{x + y}{2}\right) \cos \left(\frac{x - y}{2}\right)$

And

$\cos \left(x\right) + \cos \left(y\right) = \cos \left(a + b\right) + \cos \left(a - b\right)$

$= \cos \left(a\right) \cos \left(b\right) - \sin \left(a\right) \sin \left(b\right) + \cos \left(a\right) \cos \left(b\right) + \sin \left(a\right) \sin \left(b\right)$

$= 2 \cos \left(a\right) \cos \left(b\right)$

$= 2 \cos \left(\frac{x + y}{2}\right) \sin \left(\frac{x - y}{2}\right)$