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Math Help - Half Sine problem

  1. #1
    Member M670's Avatar
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    Half Sine problem

    Given and , find the exact value of .
    Note: You are not allowed to use decimals in your answer. = .

    \cos(\alpha)=\frac{\sqrt15}{8}=1-\sin^2(\frac{\alpha}{2})
    \sin^2(\frac{\alpha}{2})=1-\frac{\sqrt15}{8}=\frac{8-\sqrt15}{8} which then is \sin(\frac{\alpha}{2})=\sqrt\frac{8-\sqrt15}{8}

    Where did I go wrong
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  2. #2
    MHF Contributor MarkFL's Avatar
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    Re: Half Sine problem

    I think you first want to find \cos(\alpha) from:

    \cos^2(\alpha)=1-\left(\frac{7}{8} \right)^2=\frac{15}{64}

    Since \alpha is a first quadrant angle, we have:

    \cos(\alpha)=\frac{\sqrt{15}}{8}

    Next, using the double-angle identity for sine and the half-angle identity for cosine, we find:

    \frac{7}{8}=2\sqrt{\frac{1+\frac{\sqrt{15}}{8}}{2}  }\sin\left(\frac{\alpha}{2} \right)=\frac{1}{2}\sqrt{8+\sqrt{15}} \sin \left( \frac{ \alpha}{2} \right)

    Now solve for \sin\left(\frac{\alpha}{2} \right)
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  3. #3
    Member M670's Avatar
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    Re: Half Sine problem

    Can you please give me the original formulas ? so I can better understand the substitution you have made?
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  4. #4
    Member M670's Avatar
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    Re: Half Sine problem

    Should this not be \sin\frac{\alpha}{2}=\sqrt\frac{1-\cos(u)}{2} then \sin\frac{\alpha}{2}=\sqrt\frac{1-\frac{\sqrt15}{8}}{2}
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  5. #5
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    Re: Half Sine problem

    Hello, M670!'

    A couple of errors . . .


    \text{Given: }\:\sin\alpha = \tfrac{7}{8},\;\text{ and }\,0 < \alpha < \tfrac{\pi}{2}

    \text{Find the exact value of }\sin\tfrac{\alpha}{2}.

    \cos^2\!\alpha \;=\;1-\sin^2\!\alpha \;=\;1 - \left(\tfrac{7}{8}\right)^2 \;=\;1-\tfrac{49}{64} \;=\;\tfrac{15}{64}

    . . \text{Hence: }\:\cos\alpha \;=\:\frac{\sqrt{15}}{8}

    \text{Identity: }\:\sin^2\tfrac{\alpha}{2} \:=\:\frac{1 - \cos\alpha}{{\color{red}2}} \;=\;\frac{1-\frac{\sqrt{15}}{8}}{2} \;=\;\frac{8-\sqrt{15}}{16}

    \text{Therefore: }\:\sin\tfrac{\alpha}{2} \;=\;\sqrt{\frac{8-\sqrt{15}}{16}} \;=\; \frac{\sqrt{8-\sqrt{15}}}{4}
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