baselramjet
New member
- Joined
- Mar 22, 2007
- Messages
- 34
[sin(x) - sin(3x)] / [cos(x) + cos(3x)] = -tan(x)
\(\displaystyle \L\frac{\sin x\,-\,\sin 3x}{\cos x\,+\,\cos3x}\:=\:-\tan x\)
So in the example: \(\displaystyle \L\:\frac{\sin(x)\,-\,\sin(3x)}{\cos(x)\,-\,\cos(3x)}\)\(\displaystyle \: = \;-\cot(2x)\)
The sum-product identity for: \(\displaystyle \,\cos(x)\,-\,\cos(y)\)
. . would be: \(\displaystyle \,-2\cdot\sin\left(\frac{x+y}{2}\right)\cdot\sin\left(\frac{x-y}{2}\right)\;\;\) Right!
So in that example the numerator would be the same as the previous example
and the denominator would be:
\(\displaystyle \cos(x)\,-\,\cos(3x) \:=\:-2\cdot\sin\left(\frac{x+3x}{2}\right)\cdot\sin\left(\frac{x-3x}{2}\right) \:=\:-2\cdot\sin\left(\frac{4x}{2}\right)\cdot\sin\left(\frac{-2x}{2}\right)\)
. . \(\displaystyle = \:-2\cdot\sin(2x)\cdot\sin(-x)\: = \:\)+\(\displaystyle 2\cdot\sin(2x)\cdot\sin(x)\)
Am I correct? . Yes!