minimum value of $2\cos \alpha\sin \beta+3\sin \alpha\sin \beta+4\cos \beta$

Question

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

Answer 1

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

$$-\sqrt{13+4^2}.$$

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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.

Answer 2

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.

Answer 3

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 $-\sqrt{29}$.

Answer 4

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) :