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Math Help - exact value question

  1. #1
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    exact value question

    If cos (-) = 4/5 , 0 < (-) < pie/2

    find the exact value of

    (a) sin 2(-)
    (b) sin 4(-)


    given that tan x = square root of 11 all over 3, o < x < pie/2 , find the exact value of sin x



    If anyone could help me with these questions id be gratefull



    edit : damn.. can anyone help me show that sin (2x + x) = 3sin x - 4sin(to the power of 3)x
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  2. #2
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    \cos \theta = \frac{4}{5}

    using \sin^2 \theta + \cos^2 \theta = 1 or and appropriate triangle calculate the exact value of \sin \theta.

    you should now that \sin 2 \theta = 2\sin \theta \cos \theta
    this should be an easy calculation.


    similarly \sin 4 \theta = 2\sin 2 \theta \cos 2 \theta
    should be able to take it form there
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  3. #3
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    Thanks,

    i also thought i could do another question but i need a little help.

    show that sin (2x + x) = 3 sin x - 4 sin^3x
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  4. #4
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    show that  \sin (2\theta + \theta) = 3 \sin \theta - 4 \sin^3 \theta
    starting on the LHS using the compound angle formula
     \sin (2\theta + \theta) = \sin 2 \theta \cos \theta + \cos 2 \theta \sin \theta


    you know that \sin 2\theta = 2\sin \theta \cos \theta
    and you should know \cos 2\theta = cos^2 \theta -  \sin^2 \theta

    you may need to use \cos^2 \theta = 1 - \sin^2 \theta
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  5. #5
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    Quote Originally Posted by Hydralore View Post
    can anyone help me show that sin (2x + x) = 3sin x - 4sin(to the power of 3)x
    \sin 3x = \text{Im} \left( {e^{ix} } \right)^3 = \text{Im} \left( {\cos x + i\sin x} \right)^3 .

    Now

    \left( {\cos x + i\sin x} \right)^3 = \cos ^3 x + 3i\sin x\cos ^2 x - 3\sin ^2 x\cos x - i\sin ^3 x.

    And finally

    \sin 3x = 3\sin x(1 - \sin ^2 x) - \sin ^3 x = 3\sin x - 4\sin ^3 x\,\blacksquare
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  6. #6
    Senior Member JaneBennet's Avatar
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    Quote Originally Posted by Hydralore View Post
    given that tan x = square root of 11 all over 3, o < x < pie/2 , find the exact value of sin x
    Calculate the hypotenuse using Pythagorasí theorem and use \mbox{sine} = \frac{\mbox{opposite}}{\mbox{hypotenuse}}.
    Attached Thumbnails Attached Thumbnails exact value question-tanx.gif  
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  7. #7
    quq
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    Quote Originally Posted by JaneBennet View Post
    Calculate the hypotenuse using Pythagoras’ theorem and use \mbox{sine} = \frac{\mbox{opposite}}{\mbox{hypotenuse}}.


    hypotenuse = \sqrt{20}

    sin(x) = \sqrt{ \frac{11} {20} }


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