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Math Help - Trig Questions, help please :)?

  1. #1
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    Trig Questions, help please :)?

    Ok..uh, in a bit stuck.
    1. Determine the greatest possible value of
    9sin(10x / 3) - 12sin^3(10x/3)
    Well i know that the greatest value of a sin function is 1...so overall this would give - 3.
    The min value of a sin function is -1...so overall this gives the value 3..
    However the second part of the question is
    b.) determine the value of x/
    The answer is 9....i get -27...what am i doing wrong :S?
    I know that if the sin function is to be -1 then
    \frac{10x}{3} = -90
    and re-arranging gives x = -27.

    Question 2
    Solve:
    3sin(6x)cosec2x = 4
    After converting cosec x into 1/sinx and multiplying through... i get this but now im stuck lol.
    3sin(6x) - 4sin(2x) = 0

    Any help would be greatly appreciated. Thank you!
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  2. #2
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    Quote Originally Posted by AshleyT View Post
    Ok..uh, in a bit stuck.
    1. Determine the greatest possible value of
    9sin(10x / 3) - 12sin^3(10x/3)
    Well i know that the greatest value of a sin function is 1...so overall this would give - 3.
    The min value of a sin function is -1...so overall this gives the value 3..
    However the second part of the question is
    b.) determine the value of x/
    The answer is 9....i get -27...what am i doing wrong :S?
    I know that if the sin function is to be -1 then
    \frac{10x}{3} = -90
    and re-arranging gives x = -27.

    Question 2
    Solve:
    3sin(6x)cosec2x = 4
    After converting cosec x into 1/sinx and multiplying through... i get this but now im stuck lol.
    3sin(6x) - 4sin(2x) = 0

    Any help would be greatly appreciated. Thank you!
    Differentiate to get the maximum value.
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  3. #3
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    Quote Originally Posted by AshleyT View Post
    Ok..uh, in a bit stuck.
    1. Determine the greatest possible value of
    9sin(10x / 3) - 12sin^3(10x/3)

    Well i know that the greatest value of a sin function is 1...so overall this would give - 3.
    The min value of a sin function is -1...so overall this gives the value 3..

    However the second part of the question is
    b.) determine the value of x/
    The answer is 9....i get -27...what am i doing wrong :S?
    I know that if the sin function is to be -1 then
    \frac{10x}{3} = -90
    and re-arranging gives x = -27.
    You cannot simply substitute the maximum value of sin to get the maximum value of the function.

    First of all, you should know the triple angle identity: 3\sin y - 4 \sin^3 y = \sin 3y,

    This is what you had to do:

    9\sin \frac{10x}{3} - 12 \sin^3 \frac{10x}{3} = 3 \left(3\sin \frac{10x}{3} - 4 \sin^3 \frac{10x}{3}\right)

    Now use the triple angle identity,

     3 \left(3\sin \frac{10x}{3} - 4 \sin^3 \frac{10x}{3}\right) = 3 \sin 3\left(\frac{10x}{3}\right) = 3\sin 10 x

    Greatest value of 3\sin 10 x is 3 and occurs when \sin 10 x = 1

    So now you will get the answer

    Question 2
    Solve:
    3sin(6x)cosec2x = 4
    After converting cosec x into 1/sinx and multiplying through... i get this but now im stuck lol.
    3sin(6x) - 4sin(2x) = 0

    Any help would be greatly appreciated. Thank you!
    3\sin 6x - 4\sin 2x = 0

    Now use the \sin 3x = 3 \sin x - 4 \sin^3 x identity, with x replaced by 2x.

    3(3 \sin 2x - 4 \sin^3 2x ) - 4 \sin 2x = 0

    So can you continue now?
    Last edited by Isomorphism; December 29th 2008 at 08:46 AM.
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  4. #4
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    Quote Originally Posted by Isomorphism View Post
    You cannot simply substitute the maximum value of sin to get the maximum value of the function.

    First of all, you should know the triple angle identity: 3\sin y - 4 \sin^3 y = \sin 3y,

    This is what you had to do:

    9\sin \frac{10x}{3} - 12 \sin^3 \frac{10x}{3} = 3 \left(3\sin \frac{10x}{3} - 4 \sin^3 \frac{10x}{3}\right)

    Now use the triple angle identity,

     3 \left(3\sin \frac{10x}{3} - 4 \sin^3 \frac{10x}{3}\right) = 3 \sin 3\left(\frac{10x}{3}\right) = 3\sin 10 x

    Greatest value of 3\sin 10 x is 3 and occurs when \sin 10 x = 1

    So now you will get the answer

    3\sin 6x - 4\sin 2x = 0

    Now use the \sin 3x = 3 \sin x - 4 \sin^3 x identity, with x replaced by 2x.

    3(3 \sin 2x - 4 \sin^3 2x ) - 4 \sin 2x = 0

    So can you continue now?
    Thank-you so much!
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