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Math Help - Using the identity cos^2x + sin^2x = 1

  1. #1
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    Using the identity cos^2x + sin^2x = 1

    Using the identity cos^2x + sin^2x = 1, or otherwise, find an equivalent expression to 1+cosx/2sinx + sinx/2+2cosx

    If anyone is able to help. Please do.
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  2. #2
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    1+cosx/2sinx + sinx/2+2cosx

    If that is
    (1 +cosX)/(2sinX) +(sinX)/(2 +2cosX),
    then,

    Combine the two fractions into one fraction only.
    The common denominator is 2(sinX)(1+cosX),

    = [(1+cosX)^2 +(sinX)^2] / [2(sinX)(1+cosX)]
    = [1 +2cosX +cos^2(X) +sin^2(X)] / [2(sinX)(1+cosX)]
    = [1 +2cosX +1] / [2(sinX)(1+cosX)]
    = [2 +2cosX] / [(sinX)(2 +2cosX)]
    = 1 / sinX
    = cscX ------------------answer.
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  3. #3
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    Hello, haku!

    Another approach . . .


    Using the identity \cos^2\!x + \sin^2\!x \:= \:1, or otherwise,

    find an equivalent expression to: . \frac{1+\cos x}{2\sin x} + \frac{\sin x}{2+2\cos x}
    We have: . \frac{1+\cos x}{2\sin x} + \frac{\sin x}{2(1+\cos x)}

    Multiply the second fraction by: \frac{1-\cos x}{1-\cos x}

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

    The problem becomes: . \frac{1 + \cos x}{2\sin x} + \frac{1-\cos x}{2\sin x} \;=\;\frac{2}{2\sin x} \;=\;\frac{1}{\sin x} \;=\;\boxed{\csc x}

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  4. #4
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    Thanks for the help!
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