# FIND sin alpha and sin beta

• April 17th 2006, 07:08 AM
ling_c_0202
FIND sin alpha and sin beta
I do not have any idea to deal with this question, can anyone help?

It is given that
$
y = \sin (x) + 4 \sin (x)
$
,
where 0 <= x <= 2x.
If y has the maximum value when x = $\alpha$
and the minimum value when x = $\beta$,
find the values of $\sin\alpha$ and $\sin\beta$.
• April 17th 2006, 08:01 AM
ThePerfectHacker
Quote:

Originally Posted by ling_c_0202
I do not have any idea to deal with this question, can anyone help?

It is given that
$
y = \sin (x) + 4 \sin (x)
$
,
where 0 <= x <= 2pi.
If y has the maximum value when x = $\alpha$
and the minimum value when x = $\beta$,
find the values of $\sin\alpha$ and $\sin\beta$.

Thus,
$y=5\sin(x)$
The amplitude is 5, thus it max is 5. Max happens where $\sin x$ is max. Sine is max when $\sin x=1$ Thus, at $x=\pi/2$.

Similary the minimum occurs at $\sin x=-1$ thus, $x=\frac{3\pi}{2}$
• April 17th 2006, 08:17 AM
ling_c_0202
sorry
I typed the questioned wrongly.

It should be

It is given that y = sin x + 4 cos x,
where 0 < = x <= 2pi.
If y has the maximum value when x = alpha
and the minimum value when x = beta,
find the values of sin alpha and sin beta.
• April 17th 2006, 09:34 AM
ThePerfectHacker
Quote:

Originally Posted by ling_c_0202
sorry
I typed the questioned wrongly.

It should be

It is given that y = sin x + 4 cos x,
where 0 < = x <= 2pi.
If y has the maximum value when x = alpha
and the minimum value when x = beta,
find the values of sin alpha and sin beta.

Notice if $a,b>0$ and you have expression,
$a\sin x+b\cos x$, then, do the following trick,
$\sqrt{a^2+b^2}\left( \frac{a}{\sqrt{a^2+b^2}}\sin x+\frac{b}{\sqrt{a^2+b^2}}\cos x \right)$
Think, of a right triangle with legs $a,b$ then its hypotenuse is $\sqrt{a^2+b^2}$ thus, if $\theta$ is an angle between the hypotenuse and $a$ then you have,
$\left\{ \begin{array}{c}\cos \theta=\frac{a}{\sqrt{a^2+b^2}}\\ \sin \theta=\frac{b}{\sqrt{a^2+b^2}} \\ \tan \theta=\frac{b}{a} \end{array}\right$
Thus, substituting that into our expression,
$\sqrt{a^2+b^2}(\cos \theta \sin x+\sin \theta \cos x)$
Thus, sum for sine of two angles,
$\sqrt{a^2+b^2}(\sin(x+\theta))$ where $\tan^{-1} (b/a)=\theta$
------------------
Let us return back to your problem.
You have $a=1,b=4>0$ thus, using this trick you have,
$\sqrt{17}\sin(x+\theta)$ where $\theta=\tan^{-1}4$
As explained before the max occurs when,
$\sin (x+\theta)=1$
Thus,
$x+\theta=\frac{\pi}{2}$
Thus,
$x=\frac{\pi}{2}-\theta$
Thus,
$\alpha=\frac{\pi}{2}-\theta$
We need,
$\sin \alpha$
Which is,
$\sin \left( \frac{\pi}{2}-\theta \right)=\cos \theta=\cos (\tan^{-1} 4)=\frac{1}{\sqrt{17}}$
----------------------------
For the min that happens when,
$\sin(x+\theta)=-1$
Thus,
$x+\theta=\frac{3\pi}{2}$
Thus,
$x=\frac{3\pi}{2}-\theta$
Thus, $\beta=\frac{3\pi}{2}-\theta$
But you need,
$\sin \beta=\sin \left(\frac{3\pi}{2}-\theta \right)=-\cos(\theta)$
But, $-\cos(\theta)=-\frac{1}{\sqrt{17}}$
Thus,
$\sin \beta=-\frac{1}{\sqrt{17}}$