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Math Help - Trigo identities

  1. #1
    Junior Member
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    Feb 2008
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    Singapore
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    Trigo identities

    hey, ive got 2 questions here in which i dont know how to do. Im completely clueless for question 1 but i did some for question 2. hopefully someone can help me ^^ thanks!

    Q1) Prove (1 - tan A)(tan 2A + Sec 2A) = 1 + tan A
    Q2) Prove cos^4 A = \frac {1}{8} (3 + 4cos 2A + cos 4A)
    for this, i got
     \frac {3}{8} + cos^2 A - \frac{1}{2} + \frac{1}{8} cos 2A cos 2A  - \frac{1}{8} sin 2A sin 2A
    =
     cos^2 A - \frac {1}{8} + \frac {1}{8} ( 4cos^4 A - 4cos^2 A + 1) - \frac {1}{2} sin^2 A cos^2 A
    =
     cos^2 A - \frac {1}{8} + \frac {1}{2} cos^4 A - \frac{1}{2} cos^2 A + \frac{1}{8} - \frac {1}{2} sin^2 A cos^2 A
    =
     2cos^2 A - \frac {1}{4} + cos^4 A - cos^2 A + \frac {1}{4} - sin^2 A cos^2 A
    =
     cos^2 A - sin^2 A cos^2 A - \frac {1}{4} + \frac {1}{4} + cos^4 A
    and i knew that it was from this step that something is wrong, but i dont know what it is ><
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  2. #2
    MHF Contributor
    Joined
    Nov 2008
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    France
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    Quote Originally Posted by teddybear67 View Post
     cos^2 A - \frac {1}{8} + \frac {1}{2} cos^4 A - \frac{1}{2} cos^2 A + \frac{1}{8} - \frac {1}{2} sin^2 A cos^2 A
    =
     2cos^2 A - \frac {1}{4} + cos^4 A - cos^2 A + \frac {1}{4} - sin^2 A cos^2 A
    I do not understand why you multiplied everything by 2 ?

    You are close to the end :

     \cos^2 A - \frac {1}{8} + \frac {1}{2} \cos^4 A - \frac{1}{2} \cos^2 A + \frac{1}{8} - \frac {1}{2} \sin^2 A \cos^2 A = 1-\cos^2 A) \cos^2 A = \cos^4 A" alt="\frac{1}{2} \cos^2 A + \frac {1}{2} \cos^4 A - \frac {1}{2} \1-\cos^2 A) \cos^2 A = \cos^4 A" />
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