# trigo

• Aug 4th 2009, 03:20 AM
thereddevils
trigo
Given that sin(x+y) = pcos(x-y) , show that

tan x = (p-tan y)/(1-p tany)
• Aug 4th 2009, 04:49 AM
Soroban
Hello, thereddevils!

Quote:

Given that: .$\displaystyle \sin(x+y) \:=\: p\cos(x-y)$

show that: .$\displaystyle \tan x \:=\:\frac{p-\tan y}{1-p\tan y}$

We have: .$\displaystyle \sin(x+y) \;=\;p\cos(x-y)$

Then: .$\displaystyle \sin x\cos y + \cos x\sin y \;=\;p(\cos x\cos y + \sin x\sin y)$

Rearrange terms: .$\displaystyle \sin x\cos y - p\sin x\sin y \;=\;p\cos x\cos y - \cos x\sin y$

Factor: .$\displaystyle \sin x(\cos y - p\sin y) \;=\;\cos x(p\cos y - \sin y)$

. . Then: .$\displaystyle \frac{\sin x}{\cos x} \;=\;\frac{p\cos y - \sin y}{\cos y - p\sin y}$

On the right, divide numerator and denominator by $\displaystyle \cos y$:

. . . . $\displaystyle \frac{\sin x}{\cos x} \;=\;\frac{\dfrac{p\cos y}{\cos y} - \dfrac{\sin y}{\cos y}} {\dfrac{\cos y}{\cos y} - \dfrac{p\sin y}{\cos y}}$

And we have: .$\displaystyle \tan x \;=\;\frac{p-\tan y}{1 - p\tan y}$