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.
Property to note : $a\cos x + b\sin x = \pm\sqrt{a^2 +b^2}$, So,


$$2\cos\alpha + 3\sin\alpha = \pm\sqrt{13}$$ 
Taking the minimum value of the expression,
$$-\sqrt{13}\sin\beta +4\cos\beta = \pm\sqrt{29}$$
Therefore, the minimum value of the expression is $-\sqrt{29}$.