equation

**olaolaola** · November 1st 2010, 04:56 AM

We've got following equation:
sin(x+y)/sin(x) = sin(2x+y+z)/sin(z)

Find z.

**Ackbeet** · November 1st 2010, 05:04 AM

Let $M=\sin(x+y)/\sin(x),$ and let $N=2x+y.$ Then the equation becomes

$M=\sin(N+z)/\sin(z),$ or

$M\sin(z)=\sin(N+z),$ assuming $\sin(z)\not=0.$

Do you have any ideas about the RHS?

**Soroban** · November 1st 2010, 05:47 AM

Hello, olaolaola!

$\text{Solve for }z\!:\;\;\dfrac{\sin(x+y)}{\sin x} \;=\; \dfrac{\sin(2x+y+z)}{\sin z}$

We have: . $\sin(x+y)\cdot\sin z \;=\;\sin x\cdot\sin([2x+y]+z)$

. . $\sin(x+y)\cdot\sin z \;=\;\sin x\bigg[\sin(2x+y)\cdot\cos z + \cos(2x+y)\cdot\sin z\bigg]$

. . $\sin(x+y)\cdot\sin z \;=\;\sin x\cdot\sin(2x+y)\cdot\cos z + \sin x \cdot\cos(2x+y)\cdot \sin z$

. . $\sin(x+y)\cdot\sin z - \sin x\cdot\cos(2x+y)\cdot\sin z \;=\;\sin x\cdot\sin(2x+y)\cdot\cos z$

. . $\bigg[\sin(x+y)-\sin x\cdot\cos(2x+y)\bigg]\sin z \;=\;\sin x\cdot\sin(2x+y)\cdot\cos z$

. . $\displaystyle \frac{\sin z}{\cos z} \;=\;\frac{\sin x\cdot\sin(2x+y)}{\sin(x+y) - \sin x\cdot\cos(2x+y)}$

. . $\displaystyle \tan z \;=\;\frac{\sin x\cdot\sin(2x+y)}{\sin(x+y) - \sin x\cdot\cos(2x+y)}$

. . . . $z \;=\;\tan^{-1}\left[\dfrac{\sin x\cdot\sin(2x+y)}{\sin(x+y) - \sin x\cdot\cos(2x+y)}\right]$

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