# Thread: FIND sin alpha and sin beta

1. ## FIND sin alpha and sin beta

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

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

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

It is given that
$\displaystyle y = \sin (x) + 4 \sin (x)$,
where 0 <= x <= 2pi.
If y has the maximum value when x = $\displaystyle \alpha$
and the minimum value when x = $\displaystyle \beta$,
find the values of $\displaystyle \sin\alpha$ and $\displaystyle \sin\beta$.
Thus,
$\displaystyle y=5\sin(x)$
The amplitude is 5, thus it max is 5. Max happens where $\displaystyle \sin x$ is max. Sine is max when $\displaystyle \sin x=1$ Thus, at $\displaystyle x=\pi/2$.

Similary the minimum occurs at $\displaystyle \sin x=-1$ thus, $\displaystyle x=\frac{3\pi}{2}$

3. 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.

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

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