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
 A: 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.
A: Some answers mention a 2D dot product. 
But in fact one can have a more global view by interpreting the quantity to be minimized:
$$2\cos \alpha\sin \beta+3\sin \alpha\sin \beta+4\cos \beta \tag{1}$$
as the 3D dot product $U.V$ of
$$U=\begin{pmatrix}2\\3\\4\end{pmatrix} \ \ \ \text{with} \ \ \ V=\begin{pmatrix}\cos \alpha\sin \beta\\\sin \alpha\sin \beta\\\cos \beta\end{pmatrix}$$
where $V$ is a point on the unit sphere, as we recognize spherical coordinates, ($\alpha$ = longitude, $\beta$=latitude). 
Therefore, dot product $U.V$ of fixed $U$ with variable $V$ is minimal when one takes $V$ opposite to $U$ ; as $V$ is constrained to have a unit norm, this minimal dot product is  
$$U . \left(-\frac{U}{\|U\|}\right)=-\frac{U.U}{\|U\|}=-\frac{\|U\|^2}{\|U\|}=-\|U\|=-\sqrt{2^2+3^2+4^2}=-\sqrt{29}.$$
For the fun, here is a graphical representation of a little part of the doubly periodic surface with equation $z=f(\alpha,\beta)$ where the RHS is expression (1) :

A: By C-S twice we obtain: 
$$2\cos\alpha\sin\beta+3\sin\alpha\sin\beta+4\cos\beta=$$
$$=\sin\beta(2\cos\alpha+3\sin\alpha)+4\cos\beta\geq$$
$$\geq-\sqrt{(\sin^2\beta+\cos^2\beta)((2\cos\alpha+3\sin\alpha)^2+16)}\geq$$
$$\geq-\sqrt{\left(\sqrt{2^2+3^2)(\cos^2\alpha+\sin^2\alpha}\right)^2+16}=-\sqrt{29}.$$
The equality occurs for $(\cos\alpha,\sin\alpha)||(2,3)$ and $(\sin\beta,\cos\beta)||(2\cos\alpha+3\sin\alpha,4),$ 
which says that we got a minimal value.
A: 
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 $-\sqrt{29}$.
