MHB [ASK} Prove (cos2x+cos2y)/(sin2x−sin2y)=1/tan(x−y)

[Monoxdifly]

Prove that $$\frac{cos2x+cos2y}{sin2x-sin2y}=\frac1{tan(x-y)}$$. Can someone provide me some hints? I tried to manipulate the right-hand expression but got back to square one.

 

[castor28]

Prove that $$\frac{cos2x+cos2y}{sin2x-sin2y}=\frac1{tan(x-y)}$$. Can someone provide me some hints? I tried to manipulate the right-hand expression but got back to square one.

Hi Monoxdifly,

You could start with the LHS and the identities:

\begin{align*}
\cos a + \cos b &= 2\cos\frac{a+b}{2}\cos\frac{a-b}{2}\\
\sin a - \sin b &= 2\sin\frac{a-b}{2}\cos\frac{a+b}{2}
\end{align*}

 

[Monoxdifly]

Hi Monoxdifly,

You could start with the LHS and the identities:

\begin{align*}
\cos a + \cos b &= 2\cos\frac{a+b}{2}\cos\frac{a-b}{2}\\
\sin a - \sin b &= 2\sin\frac{a-b}{2}\cos\frac{a+b}{2}
\end{align*}

Ah, let's see...
$$\frac{cos2x+cos2y}{sin2x-sin2y}$$=$$\frac{2cos\frac{2x+2y}{2}cos\frac{2x-2y}{2}}{2sin\frac{2x-2y}{2}cos\frac{2x+2y}{2}}$$=$$\frac{cos(x-y)}{sin(x-y)}$$= cot(x - y) = $$\frac1{tan(x-y)}$$
Wew. Just 4 steps.