1. ## Trigonomic Identity Problem

Hello

I am new at this forum, but I was wondering if I could get some help.

The identity problem I am trying to solve is:

sinx-sin3x
---------- = -cot2x
cosx-cos3x

(note, the sin3x, cos3x, and cot2x do not mean to the square root of)

Thanks for any help with this

2. Hello, themartinacster!

Welcome aboard!

There are two ways (at least) to prove the identity,
. . neither is pleasant.

Prove: . $\frac{\sin x - \sin3x}{\cos x - \cos 3x} \:=\:-\cot2x$

[1] Use the Sum-to-Product identities:

. . . $\begin{array}{ccc}\sin A - \sin B &=& 2\cos\left(\frac{A+B}{2}\right)\sin\left(\frac{A-B}{2}\right) \\ \\[-3mm]
\cos A - \cos B &=& \text{-}2\sin\left(\frac{A+B}{2}\right)\sin\left(\frac{A-B}{2}\right) \end{array}$

We have: . $\begin{array}{ccccc}\sin x - \sin3x &=& 2\cos2x\sin(-x) &=&\text{-}2\cos2x\sin x \\

\cos3x - \cos x &=& \text{-}2\sin2x\sin(-x) &=& 2\sin2x\sin x \end{array}$

The problem becomes: . $\frac{\sin x-\sin3x}{\cos x - \cos3x} \;=\;\frac{\text{-}2\cos2x\sin x}{2\sin2x\sin x} \;=\;-\frac{\cos2x}{\sin2x} \;=\;-\cot2x$

~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~

[2] Use the Compound Angle identities:

. . . $\begin{array}{ccc}\sin(A + B) &=& \sin A\cos B + \sin B\cos A \\ \cos(A + B) &=& \cos A\cos B - \sin A\sin B \end{array}$

Use them repeatedly to derive these statements:

. . $\begin{array}{ccc}\sin3x &=& 3\sin x - 4\sin^3\!x \\ \cos3x &=& 4\cos^3\!x - 3\cos x \end{array}$

We have: . $\frac{\sin x - \sin3x}{\cos x - \cos3x} \;=\;\frac{\sin x - (3\sin x - 4\sin^3\!x)}{\cos x - (4\cos^3\!x - 3\cos x)} \;=\;\frac{4\sin^3\!x - 2\sin x}{4\cos x-4\cos^3\!x}$

. . . $= \;\frac{2\sin x(2\sin^2\!x - 1)}{4\cos x\underbrace{(1 - \cos^2x)}_{\text{This is }\sin^2\!x}} \;=\;-\frac{\sin x\overbrace{(1 - 2\sin^2\!x)}^{\text{This is }\cos2x}}{2\cos x\sin^2\!x}$

. . . $= \;-\frac{\cos2x}{\underbrace{2\sin x\cos x}_{\text{This is }\sin2x}} \;=\;-\frac{\cos2x}{\sin2x} \;=\;-\cot2x$