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Math Help - Simple Trig Help #2

  1. #1
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    Simple Trig Help #2

    Can someone check this:

    (\cos60-\tan45)(\sin30+\tan45)

    (\frac{\sqrt3}{2} - 1)(\frac {\sqrt3}{2} +1)

    \frac{3}{4} -1

    \frac {-3}{4}

    Thanks in advance.
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  2. #2
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    Quote Originally Posted by jgv115 View Post
    Can someone check this:

    (\cos60-\tan45)(\sin30+\tan45)

    (\frac{\sqrt3}{2} - 1)(\frac {\sqrt3}{2} +1)

    \frac{3}{4} -1

    \frac {-3}{4}

    Thanks in advance.
    \sin{30^{\circ}} = \frac{1}{2} and \cos{60^{\circ}} = \frac{1}{2}

    So (\cos{60^\circ} - \tan{45^\circ})(\sin{30^\circ} + \tan{45^\circ}) = \left(\frac{1}{2} - 1\right)\left(\frac{1}{2} + 1\right)

     = \left(-\frac{1}{2}\right)\left(\frac{3}{2}\right)

     = -\frac{3}{4}.
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    Simplify

    \frac {2\csc^2 \theta-2}{4\cot\theta}

    Well I just factorised the top bit:

     \frac {2(\csc^2\theta-1)}{4\cot\theta}

    I don't know where to go from here. Any help will be appreciated.
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  4. #4
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    Quote Originally Posted by jgv115 View Post
    Simplify

    \frac {2\csc^2 \theta-2}{4\cot\theta}

    Well I just factorised the top bit:

     \frac {2(\csc^2\theta-1)}{4\cot\theta}

    I don't know where to go from here. Any help will be appreciated.
    Think about the Pythagorean Identity:

    \sin^2{\theta} + \cos^2{\theta} = 1.

    Divide both sides by \sin^2{\theta} and you get

    1 + \cot^2{\theta} = \csc^2{\theta}.


    So what do you think \csc^2{\theta} - 1 equals?
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