minimum value of $2\cos \alpha\sin \beta+3\sin \alpha\sin \beta+4\cos \beta$ Let $\alpha,\beta$ be real numbers ; find the minimum value of 
$2\cos \alpha\sin \beta+3\sin \alpha\sin \beta+4\cos \beta$
What I tried :
$\bigg|4\cos \beta+(2\cos \alpha+3\sin \alpha)\sin \beta\bigg|^2 \leq 4^2+(2\cos \alpha+3\sin \alpha)^2$
How do I solve it ? Help me please

In 
$$(2\cos\alpha+3\sin\alpha)\sin\beta+4\cos\beta$$ the parenthesised factor takes values in $[-\sqrt{2^2+3^2},\sqrt{2^2+3^2}]$. Using the largest value, the minimum of $$\sqrt13\sin\beta+4\cos\beta$$ is
$$-\sqrt{13+4^2}.$$

Justification:
$$a\cos t+b\sin t$$ is the scalar product of $(a,b)$ with the unit rotating vector $(\cos\theta,\sin\theta)$, which takes the extreme values $\pm\|(a,b)\|=\pm\sqrt{a^2+b^2}$.
We use this property twice.
