# Thread: another question!

1. ## another question!

sin(x)/sin(y)= 1/2 cos(x)/cos(y)= 3

prove sin(x+y) = (7/3)sin(x)cos(x)

I dont know what to do with the fractions, do i solve for one and then try and put it in the equation or something?

like sin(x)= (1/2)sin(y)

2. Originally Posted by OmegaCenturion
sin(x)/sin(y)= 1/2 cos(x)/cos(y)= 3

prove sin(x+y) = (7/3)sin(x)cos(x)

I dont know what to do with the fractions, do i solve for one and then try and put it in the equation or something?
Um, yes I think so.

like sin(x)= (1/2)sin(y)
Yes: Basically, the given (two) equations allow you to express sin(x) as a multiple of sin(y), and cos(x) as a multiple of cos(y). Thus, by applying the addition formula to sin(x+y), i.e. sin(x+y)=sin(x) cos(y)+cos(x) sin(y), you can reduce both sides of the equality that you have to prove to a certain multiple of sin(y) and cos(y). Therefore you only have to check whether, in fact, both sides work out to be the exact same multiple of sin(y) and cos(y).

3. Hello, OmegaCenturion!

Given: .$\displaystyle \begin{Bmatrix}\dfrac{\sin x}{\sin y} &=& \frac{1}{2} & [1] \\ \\[-3mm] \dfrac{\cos x}{\cos y}&=& 3 & [2] \end{Bmatrix}$

Prove: .$\displaystyle \sin(x+y) \:=\: \tfrac{7}{3}\sin x \cos x$

From [1] we have: .$\displaystyle \sin y \:=\:2\sin x$ .(a)

From [2] we have: .$\displaystyle \cos y \:=\:\tfrac{1}{3}\cos x$ .(b)

We know that: .$\displaystyle \sin(x+y) \;=\;\sin x\cos y + \cos x\sin y$ .(c)

Substitute (a) and (b) into (c):

. . $\displaystyle \sin(x+y) \;=\;\sin x\left(\tfrac{1}{3}\cos x\right) + \cos x\left(2\sin x\right)$

. . . . . . . . .$\displaystyle =\; \tfrac{1}{3}\sin x\cos x + 2\sin x\cos x$

. . . . . . . . .$\displaystyle =\;\tfrac{7}{3}\sin x\cos x$