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Math Help - Prove that

  1. #1
    Newbie Sanjana Das's Avatar
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    Question Prove that


    If Tn = sin^nA+cos^nA
    Prove that
    T3-T5 = T5-T7
    ...T1.......T3


    plzzzz explain by covering all the steps as i belong to 10th standard ......
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  2. #2
    Super Member malaygoel's Avatar
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    Quote Originally Posted by Sanjana Das View Post
    If Tn = sin^nA+cos^nA
    Prove that
    T3-T5 = T5-T7
    ...T1.......T3


    plzzzz explain by covering all the steps as i belong to 10th standard ......
    = \frac{T_5-T_7}{T_3-T_5}

    \frac{sin^5A+cos^5A-sin^7A-cos^7A}{sin^3A+cos^3A-sin^5A-cos^5A}

    = \frac{sin^5A-sin^7A+cos^5Asin^2A}{sin^3A-sin^5A+cos^3Asin^2A}

    = \frac{sin^3A-sin^5A+cos^5A}{sinA-sin^3A+cos^3A}

    = \frac{sin^3Acos^2A+cos^5A}{sinAcos^2A+cos^3A}

    = \frac{sin^3A+cos^3A}{sinA+cosA}

    = \frac{T_3}{T_1}
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  3. #3
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    (T3 - T5) = sin^3A + cos^3A -sin^5A - cos^5A
    = sin^3A - sin^5A + cos^3A - cos^5A
    = sin^3A( 1 - sin^2A) + cos^3A ( 1- cos^2A)
    = sin^3A*cos^2A + cos^3A*sin^2A
    = sin^2A*cos^2A(sinA + cosA)
    If you divide this by T1 you get sin^2A*cos^2A.
    Proceed in the same manner in right hand side.
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  4. #4
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    He3llo, Sanjana Das!

    Another approach . . .


    If T_n \:=\:\sin^n\!A+\cos^n\!A

    prove that: . \frac{T_3-T_5}{T_1}\:=\:\frac{T_5 - T_7}{T_3}

    Left side:

    . . \frac{T_3 - T_5}{T_1} \;=\;\frac{(\sin^3\!A + \cos^3\!A) - (\sin^5\!A + \cos^5\!A)}{\sin A + \cos A} \;= \frac{(\sin^3\!A - \sin^5\!A) - (\cos^3\!A - \cos^5\!A)}{\sin A + \cos A}

    . . . . . . . = \;\frac{\sin^3\!A(1 - \sin^2\!A) + \cos^3\!A(1 - \cos^2\!A)}{\sin A+\cos A} \;=\;\frac{\sin^3\!A\cos^2\!A + \cos^3\!A\sin^2\!A}{\sin A+\cos A}

    . . . . . . . =\;\frac{\sin^2\!A\cos^2\!A\cdot(\sin A + \cos A)}{\sin A + \cos A} \;=\;\sin^2\!A\cos^2\!A


    Right side:

    . . \frac{T_5-T_7}{T_3} \;=\;\frac{(\sin^5\!A+\cos^5\!A) - (\sin^7\!A + \cos^7\!A)}{\sin^3\!A + \cos^3\!A} \;= \;\frac{(\sin^5\! - \sin^7\!A) + (\cos^5\!A - \cos^7\!A)}{\sin^3\!A + \cos^3\!A}

    . . . . . . . = \;\frac{\sin^5\!A(1 - \sin^2\!A) + \cos^5\!A( 1 - \cos^2\!A)}{\sin^3\!A + \cos^3\!A} \;= \;\frac{\sin^5\!A\cos^2\!A + \cos^5\!A\sin^2\!A}{\sin^3\!A + \cos^3\!A}

    . . . . . . . = \;\frac{\sin^2\!A\cos^2\!A\cdot(\sin^3\!A + \cos^3\!A)}{\sin^3\!A + \cos^3\!A} \;=\;\sin^2\!A\cos^2\!A



    Therefore: . \frac{T_3-T_5}{T_1} \;=\;\frac{T_5-T_7}{T_3}



    Edit: Too slow again . . . Sa-ri-ga-ma beat me to it!
    .
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  5. #5
    Super Member malaygoel's Avatar
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    Noticed this:




    . . \frac{T_n - T_{n+2}}{T_{n-2}} \;=\;\frac{(\sin^n\!A + \cos^n\!A) - (\sin^{n+2}\!A + \cos^{n+2}\!A)}{\sin^{n-2}\!A + \cos^{n-2}\!A} \;= \frac{(\sin^n\!A - \sin^{n+2}\!A) - (\cos^n\!A - \cos^{n+2}\!A)}{\sin^{n-2}\!A + \cos^{n-2}\!A}

    . . . . . . . = \;\frac{\sin^n\!A(1 - \sin^2\!A) + \cos^n\!A(1 - \cos^2\!A)}{\sin^{n-2}\!A + \cos^{n-2}\!A} \;=\;\frac{\sin^n\!A\cos^2\!A + \cos^n\!A\sin^2\!A}{\sin^{n-2}\!A + \cos^{n-2}\!A}

    . . . . . . . =\;\frac{\sin^2\!A\cos^2\!A\cdot(\sin^{n-2}\!A + \cos^{n-2}\!A)}{\sin^{n-2}\!A + \cos^{n-2}\!A} \;=\;\sin^2\!A\cos^2\!A
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