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#### Monoxdifly

##### Well-known member

- Aug 6, 2015

- 267

- Thread starter Monoxdifly
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- Thread starter
- #1

- Aug 6, 2015

- 267

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*}

- Thread starter
- #3

- Aug 6, 2015

- 267

Ah, let's see...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*}

\(\displaystyle \frac{cos2x+cos2y}{sin2x-sin2y}\)=\(\displaystyle \frac{2cos\frac{2x+2y}{2}cos\frac{2x-2y}{2}}{2sin\frac{2x-2y}{2}cos\frac{2x+2y}{2}}\)=\(\displaystyle \frac{cos(x-y)}{sin(x-y)}\)= cot(x - y) = \(\displaystyle \frac1{tan(x-y)}\)

Wew. Just 4 steps.

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