3
$\begingroup$

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

$\endgroup$
  • $\begingroup$ is the answer -9 $\endgroup$ – rash Mar 11 at 11:15
  • 1
    $\begingroup$ Cancel the gradient. $\endgroup$ – Yves Daoust Mar 11 at 11:15
  • 1
    $\begingroup$ @rash: this is not possible as it would require all factors to be $-1$ simultaneously. $\endgroup$ – Yves Daoust Mar 11 at 11:21
3
$\begingroup$

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.

$\endgroup$
3
$\begingroup$

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) :

enter image description here

$\endgroup$
1
$\begingroup$

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.

$\endgroup$
  • $\begingroup$ I propose an answer which takes into account the fact that one can recognize spherical coordinates. $\endgroup$ – Jean Marie Mar 11 at 12:50
  • $\begingroup$ Yes, I saw. Nice! $\endgroup$ – Michael Rozenberg Mar 11 at 13:05
  • $\begingroup$ 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 $a\cos t +b\sin t=r (\sin u \cos t+\cos u \sin t)=r\sin (u+t).$ $\endgroup$ – DanielWainfleet Mar 13 at 3:04
-1
$\begingroup$

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 $-\sqrt{29}$.

$\endgroup$
  • 2
    $\begingroup$ 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. $\endgroup$ – Jean Marie Mar 11 at 12:27
  • 1
    $\begingroup$ @JeanMarie What do you mean? It is an actual property of equations in the form $a\cos x + b\sin x$ $\endgroup$ – rash Mar 11 at 13:11
  • 1
    $\begingroup$ 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 $-\sqrt{a^2+b^2}\leq a \cos(x)+b \sin(x) \leq \sqrt{a^2+b^2}$... $\endgroup$ – Jean Marie Mar 11 at 13:24
  • $\begingroup$ @rash: no, take $x=0$, and $a\ne\pm\sqrt{a^2+b^2}$. $\endgroup$ – Yves Daoust Mar 11 at 13:58
  • $\begingroup$ 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 ? $\endgroup$ – Jean Marie Mar 11 at 21:31

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.