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Math Help - 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
    <br />
y = \sin (x) + 4 \sin (x)<br />
,
    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.
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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
    <br />
y = \sin (x) + 4 \sin (x)<br />
,
    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}
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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 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}}
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