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Math Help - trigo > i cant do it!

  1. #1
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    Post trigo > i cant do it!

    Find the required expression without using a calculator :
    1. sin2A, give sinA=3/5
    2. cos2A, given cosA=3/5
    3. tan2A, given tanA=1/2

    Now my problem is use the trigo identities to find sin2A. cos2A and tan2A. i know can use triangle to find, but problem is not allow use triangle to find exact value.



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  2. #2
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    Quote Originally Posted by sanikui View Post
    Find the required expression without using a calculator :
    1. sin2A, give sinA=3/5
    2. cos2A, given cosA=3/5
    3. tan2A, given tanA=1/2

    Now my problem is use the trigo identities to find sin2A. cos2A and tan2A. i know can use triangle to find, but problem is not allow use triangle to find exact value.



    THX who Helping!
    Are you familiar with the double angle formulae?

    1. sin A = 3/5 => cos A = 4/5 (assuming 1st quadrant).

    sin (2A) = 2 sin A cos A = ....

    2. Done in a similar way to 1. Use the formula for cos (2A).

    3. Substitute tan A = 1/2 into the formula for tan (2A).
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  3. #3
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    Like this question...
    Find, without the use of a calculator, the value of sinA, given that cos2A=5/13

    Solution:
    cos(2A)=1-2sin^(2A)
    2sin^(2A)=1-cos(2A)
    sin^(2A)=1/2(1-cos(2A)) =1/2(1-5/13)
    sin(A)=2 square root 13/13
    Last edited by sanikui; September 9th 2008 at 05:47 AM.
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  4. #4
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    Quote Originally Posted by sanikui View Post
    Like this question...
    Find, without the use of a calculator, the value of sinA, given that cos2A=5/13

    Solution:
    cos2A=1-2sin^2A
    2sin^2A=1-cos2A
    sin^2A=1/2(1-cos2A) =1/2(1-5/13)
    sinA=2 square root 13/13
    So what's the problem? The only thing I'd add is a justification for why the negative solution for sin A is rejected.
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