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Thread: FIND sin alpha and sin beta

  1. #1
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    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 $\displaystyle \sin\beta$.
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  2. #2
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    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
    $\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, $\displaystyle x=\frac{3\pi}{2}$
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  3. #3
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    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.
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  4. #4
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    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 $\displaystyle a,b>0$ 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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