# 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

• is the answer -9 – rash Mar 11 at 11:15
• Cancel the gradient. – Yves Daoust Mar 11 at 11:15
• @rash: this is not possible as it would require all factors to be $-1$ simultaneously. – Yves Daoust Mar 11 at 11:21

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.

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) : 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.

• I propose an answer which takes into account the fact that one can recognize spherical coordinates. – Jean Marie Mar 11 at 12:50
• Yes, I saw. Nice! – Michael Rozenberg Mar 11 at 13:05
• Let $r=\sqrt {a^2+b^2}$ . If $a,b$ are not both $0,$ there exists $u$ such that $\sin u=a/r$ and $\cos u=b/r,$ so $a\cos t +b\sin t=r (\sin u \cos t+\cos u \sin t)=r\sin (u+t).$ – DanielWainfleet Mar 13 at 3:04

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 $$-\sqrt{29}$$.

• When you say "Property to note : $a\cos x + b\sin x = \pm\sqrt{a^2 +b^2}$" you surely mean something else, but as it is written, it is meaningless. – Jean Marie Mar 11 at 12:27
• @JeanMarie What do you mean? It is an actual property of equations in the form $a\cos x + b\sin x$ – rash Mar 11 at 13:11
• You have an equation with $x$ on the left and no $x$ on the right. You should have written $a \cos(x)+b \sin(x)=\sqrt{a^2+b^2}\cos(x-\alpha)$ for some $\alpha$, or a double inequality $-\sqrt{a^2+b^2}\leq a \cos(x)+b \sin(x) \leq \sqrt{a^2+b^2}$... – Jean Marie Mar 11 at 13:24
• @rash: no, take $x=0$, and $a\ne\pm\sqrt{a^2+b^2}$. – Yves Daoust Mar 11 at 13:58
• It is rather surprising that you do not correct at least your first sentence. Maybe you are a (good) student in a secondary school and you haven't be taught how to write a mathematical text ? – Jean Marie Mar 11 at 21:31