1. ## Simultaneous equations

cos(x)-sin(x-y) = 0
-sin(y) + sin(x-y)=0

however the solutions must be between 0 and pi/2, after a long while messing with it i got y=pi/6 and x=pi/3...can anyone show me how to actually calculate this thought, as obviously my way is pretty pointless.

2. Hello, pkr!

This is a tricky one, but I found a way . . .

$\begin{array}{cccc}{\color{blue}(1)} & \cos(x)-\sin(x-y) &=& 0 \\ {\color{blue}(2)} & \text{-}\sin(y) + \sin(x-y)&=&0 \end{array} \qquad 0 \,\leq \,x,y \,\leq \frac{\pi}{2}$

We have: . $\begin{array}{ccccc} \cos x &=& \sin(x-y) \\ \sin y &=& \sin(x-y)\end{array} \quad\Rightarrow\quad \cos x \:=\:\sin y$

. . Hence, $x\text{ and }y$ are complementary: . $y \:=\:\tfrac{\pi}{2} - x$ .[3]

Substitute into (1): . $\cos x - \sin\left(x - [\tfrac{\pi}{2} - x]\right) \:=\:0 \quad\Rightarrow\quad\cos x - \sin(2x-\tfrac{\pi}{2}) \:=\:0$

. . $\cos x - \bigg[\sin2x\cos\tfrac{\pi}{2} - \sin\tfrac{\pi}{2}\cos2x\bigg] \:=\:0 \quad\Rightarrow\quad\cos x + \cos2x \:=\:0$

. . $\cos x + 2\cos^2\!x - 1 \:=\:0\quad\Rightarrow\quad 2\cos^2\!x + \cos x - 1 \:=\:0$

This factors: . $(\cos x + 1)(2\cos x - 1) \:=\:0$

And we have: . $\cos x + 1 \;=\:0 \quad\Rightarrow\quad\cos x \:=\:-1 \quad\Rightarrow\quad{\color{red}\rlap{//////}} x \:=\:\pi$

. . and: . $2\cos x - 1\:=\:0 \quad\Rightarrow\quad \cos x \:=\:\tfrac{1}{2} \quad\Rightarrow\quad \boxed{x\:=\:\frac{\pi}{3}}$

Substitute into [3]: . $y \:=\:\tfrac{\pi}{2} - \tfrac{\pi}{3} \quad\Rightarrow\quad \boxed{y\:=\:\frac{\pi}{6}}$