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  1. #1
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    Help!

    Use the formula for cos(A+B) to show that cos^2x=\frac{1}{2}(1+cos2x), and write down a similar expression for sin^2x.

    Given that f(x) = 10cos^2x+2sin^2x=6sinxcosx, express f(x) in the form Q + Rsin(2x+a), where Q, R are constants tana=\frac{4}{3}.

    Hence find the greatest and least values of  f(x).

    I could solve the first part using other formulas but not cos(A+B) formula...

    Having problems with expressing sin^2x in a similar way...

    Also asking if if the first paragraph is related to the second paragraph of questions...
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  2. #2
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    Quote Originally Posted by Punch View Post
    Use the formula for cos(A+B) to show that cos^2x=\frac{1}{2}(1+cos2x), and write down a similar expression for sin^2x.

    Given that f(x) = 10cos^2x+2sin^2x=6sinxcosx, express f(x) in the form Q + Rsin(2x+a), where Q, R are constants tana=\frac{4}{3}.

    Hence find the greatest and least values of  f(x).

    I could solve the first part using other formulas but not cos(A+B) formula...

    Having problems with expressing sin^2x in a similar way...

    Also asking if if the first paragraph is related to the second paragraph of questions...
    1. \cos{(\alpha + \beta)} = \cos{\alpha}\cos{\beta} - \sin{\alpha}\sin{\beta}.


    So if \alpha = \beta = x, then

    \cos{(x + x)} = \cos{x}\cos{x} - \sin{x}\sin{x}

    \cos{2x} = \cos^2{x} - \sin^2{x}



    Solving for \cos^2{x}...

    \cos{2x} = \cos^2{x} - \sin^2{x}

    \cos{2x} = \cos^2{x} - (1 - \cos^2{x})

    \cos{2x} = 2\cos^2{x} - 1

    1 + \cos{2x} = 2\cos^2{x}

    \cos^2{x} = \frac{1}{2}(1 + \cos{2x}).



    Solving for \sin^2{x}...

    \cos{2x} = \cos^2{x} - \sin^2{x}

    \cos{2x} = 1 - \sin^2{x} - \sin^2{x}

    \cos{2x} = 1 - 2\sin^2{x}

    2\sin^2{x} = 1 - \cos{2x}

    \sin^2{x} = \frac{1}{2}(1 - \cos{2x}).
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