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Math Help - How to prove?trigo

  1. #1
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    Post How to prove?trigo

    1. 1+cos2A+cosA=cotA
    sin2A+sinA

    2. sin2A = tanA
    1+cos2A

    3. sin2A = cotA
    1-cos2A

    4. sin2A-cos2A=secA
    sinA cosA

    5. 1+sinA-cosA =tan A
    1+sinA+cosA 2




    THANK YOU!!
    Last edited by sanikui; September 4th 2008 at 05:26 PM.
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  2. #2
    Like a stone-audioslave ADARSH's Avatar
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    1st question!

    Hello !
    cos(2A)=cos^2(A)-sin^2(A)
    sin(2A)=2sin(A)cos(A)
    put in the above to get
    [2cos^2 (A)+cos(A)] / [2sin(A)cos(A)+sin(A)]
    Take cos(A) from numerator
    & sin(A) from denominatorcommon you get cot(A)
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  3. #3
    Like a stone-audioslave ADARSH's Avatar
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    2nd & 3rd

    put
    1+cos(2A)=2cos^2(A) for 2nd denom.

    1-cos(2A)=2sin^2(A) for 3rd denom.


    sin(2A)=2sin(A)cos(A) for numerator

    and cancel common terms.
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  4. #4
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    k...i got it....thx
    how about 4 n 5?
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  5. #5
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    Hello, sanikui!

    Here's #4 . . .


    4)\;\;\frac{\sin2A}{\sin A} - \frac{\cos2A}{\cos A} \:=\:\sec A

    We have: . \frac{\overbrace{2\sin A\cos A}^{\sin2A}}{\sin A} - \frac{\overbrace{2\cos^2\!A-1}^{\cos2A}}{\cos x} \;\;=\;\;\frac{2\sin A\cos A}{\sin A} - \frac{2\cos^2A}{\cos A} + \frac{1}{\cos A}

    . . = \;\;2\cos A - 2\cos A + \frac{1}{\cos A}\;\;=\;\;\frac{1}{\cos A}\;\;=\;\;\sec A

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  6. #6
    Like a stone-audioslave ADARSH's Avatar
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    5th one

    multiply denominator with both numer. and denom.
    u will get
    RHS=
    [1+1+2sin(A)-2cos(A) -2cos(A)sin(A)] / [1+sin^2(A)+2sin(A)-cos^2(A)]
    numerator =[2(1+sin(A))(1-cos(A)]denominator =[2sin(A)(1+sin(A)]
    cancel 2(1+sin(A)
    =>RHS = [1-cos(A)]   / [sin(A)]
    Now u remember 3rd Question
    u will get 1/cot(A/2)= tan(A/2)

    dere may be shortcut for this!
    sOrRy for I went offline ys'trday after 3rd
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  7. #7
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    Thx ADARSH and Soroban,i try my best~!
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  8. #8
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    Question 5
    i think no nid multiply denominator with both numer and denom.
    Just use half-angles formulas...then can do it.
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