Under what conditions is $\sin x\sin(x+y)$ positive for $x \in [0,\pi/2]$ and $y \in [0,\pi]$? 
Given the following relation between two angles $x$ and $y$:
$$f=\sin x\sin(x+y)$$
so that $x \in [0,\pi/2]$ and $y \in [0,\pi]$. What are the conditions so that this function is always positive?

It has two possible ways, either both $sines$ are positive or both negative. I got lost how to do that. I tried to simplify this function using:
$$\sin a\sin b=\frac{1}{2}[\cos(a-b)-\cos(a+b)]$$
which means we need to think about that sign of $\cos x - \cos(2x+y)$. Any suggestions?

All of tangent, sine and cosine are non-negative from $[0,\pi/2]$ then only sine, only tangent, and only cosine in the next three quadrants, so sine is non-negative for $x$ in $[0,\pi]$.
Didn't your teacher give you some mnemonic to remember "All, Sine, Tangent, Cosine"?

$$f=\sin x\sin(x+y)$$


so that $x \in [0,\pi/2]$ and $y \in [0,\pi]$

Then $\sin x$ is always non-negative and $\sin(x+y)$ is positive as long as $0 < x+y <\pi$. It is zero at $\pi$ and then negative for $\pi < x \leq \frac{3\pi}{2}$. Therefore, the desired conditions are $x \in (0, \frac{\pi}{2})$ and $x + y \in (0, \pi)$.
