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Thread: Simple Trig Help #2

  1. #1
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    Simple Trig Help #2

    Can someone check this:

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

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

    $\displaystyle \frac{3}{4} -1$

    $\displaystyle \frac {-3}{4}$

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

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

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

    $\displaystyle \frac{3}{4} -1$

    $\displaystyle \frac {-3}{4}$

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

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

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

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

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

    Well I just factorised the top bit:

    $\displaystyle \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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    Quote Originally Posted by jgv115 View Post
    Simplify

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

    Well I just factorised the top bit:

    $\displaystyle \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:

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

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

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


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