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Trigonometry

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determine real numbers a and b so that the expression 8sin^2(theta) + 2cos^2(theta) can be rewritten as asin^2(theta) + b

I just need help in how to start it.

  • Trigonometry - ,

    8sin^2+2cos^2=8sin^2+2(1-sin^2)=>
    a=6 & b=2

  • Trigonometry - ,

    8 sin^2(theta) + 2 cos^2(theta) = 6 sin^2(theta) + 2 [ sin^2(theta) + 2 cos^2(theta) ] =

    6 sin^2(theta) + 2 [ sin^2(theta) + cos^2(theta) ]


    Remark:

    sin^2(theta) + cos^2(theta) = 1



    6 sin^2(theta) + 2 [ sin^2(theta) + cos^2(theta) ] = 6 sin^2(theta) + 2 * 1 =

    6 sin^2(theta) + 2


    8 sin^2(theta) + 2 cos^2(theta) = 6 sin^2(theta) + 2 =


    6 sin^2(theta) + 2 = a sin^2(theta) + b


    Obviously:


    a = 6

    b = 2

  • Trigonometry - ,

    8 sin^2(theta) + 2 cos^2(theta) = 6 sin^2(theta) + 2 sin^2(theta) + 2 cos^2(theta) = 6 sin^2(theta) + 2 [ sin^2(theta) + cos^2(theta) ]

  • Trigonometry - ,

    thanks

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