The equation $$\frac{cos2x+cos2y}{sin2x-sin2y}=\frac1{tan(x-y)}$$ can be proven using trigonometric identities. By applying the sum-to-product identities, specifically $$\cos a + \cos b = 2\cos\frac{a+b}{2}\cos\frac{a-b}{2}$$ and $$\sin a - \sin b = 2\sin\frac{a-b}{2}\cos\frac{a+b}{2}$$, the left-hand side simplifies to $$\frac{cos(x-y)}{sin(x-y)}$$, which equals $$cot(x - y)$$, confirming the original equation.
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Monoxdifly said: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 said: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*}