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Thread: Need help with 2 Trigonometric Questions

  1. #1
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    Need help with 2 Trigonometric Questions

    1) if , evaluate
    a) cos x
    b) sinx/4


    2 if , evaluate

    a) sin 2x
    b) tan 2x


    if possible, plz also explain all steps

    Thanks
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  2. #2
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    Hello, supersaiyan!

    1) If $\displaystyle \sin\left(\frac{x}{2}\right) = \frac{2}{3},\;\;0 \leq x \leq \frac{\pi}{2}$

    evaluate: .$\displaystyle (a)\;\cos x\qquad (b)\;\sin\left(\frac{x}{4}\right)$
    (a) We're told that: .$\displaystyle \sin\left(\frac{x}{2}\right) \:=\:\frac{2}{3} \:=\:\frac{opp}{hyp}$

    We have a right triangle with angle $\displaystyle \frac{x}{2}$ and: $\displaystyle opp = 2,\;hyp = 3$
    Using Pythagorus, we find that: .$\displaystyle adj = \pm\sqrt{5}$
    Since $\displaystyle x$ is in Quadrant 1, $\displaystyle \frac{x}{2}$ is in Quadrant 1.
    . . Hence: .$\displaystyle \cos\left(\frac{x}{2}\right) \:=\:\frac{\sqrt{5}}{3}$


    Double-angle Identity: .$\displaystyle \cos2\theta \:=\:\cos^2\!\theta - \sin^2\!\theta$

    Hence: .$\displaystyle \cos x \;\;=\;\;\cos^2\!\left(\frac{x}{2}\right) - \sin^2\!\left(\frac{x}{2}\right)\;\;=\;\;\left(\fr ac{\sqrt{5}}{3}\right)^2 - \left(\frac{2}{3}\right)^2 \;\;=\;\;\frac{5}{9} - \frac{4}{9} \;\;=\;\;\boxed{\frac{1}{9}}$


    (b) Half-angle Identity: .$\displaystyle \sin\frac{\theta}{2} \;=\;\sqrt{\frac{1-\cos\theta}{2}} $

    $\displaystyle \text{Therefore: }\;\sin\left(\frac{x}{4}\right) \;=\;\sqrt{\frac{1-\cos\left(\frac{x}{2}\right)}{2}} \;=\;\sqrt{\frac{1-\frac{\sqrt{5}}{3}}{2}} \;=\;\boxed{\sqrt{\frac{3-\sqrt{5}}{6}}}$




    2. If $\displaystyle \tan x = -\frac{3}{4},\;\;\frac{\pi}{2} \leq x \leq \pi$

    evaluate: .$\displaystyle a)\;\sin 2x\qquad (b)\;\tan 2x$
    We are given: .$\displaystyle \tan x \:=\:-\frac{3}{4} \:=\:\frac{opp}{adj}$

    Since $\displaystyle x$ is in Quadrant 2: .$\displaystyle opp = 3,\;adj = -4$ . . . Then: .$\displaystyle hyp = 5$
    Hence: .$\displaystyle \sin x = \frac{3}{5},\;\cos x = -\frac{4}{5}$


    $\displaystyle (a)\;\sin2x \;=\;2\sin x\cos x \;=\;2\left(\frac{3}{5}\right)\left(-\frac{4}{5}\right) \;=\;\boxed{-\frac{24}{27}}$

    $\displaystyle (b)\;\tan2x \;=\;\frac{2\tan x}{1-\tan^2\!x} \;=\;\frac{2\left(-\frac{3}{4}\right)}{1 - \left(-\frac{3}{4}\right)^2} \;=\;\frac{-\frac{3}{2}}{\frac{7}{16}} \;=\;\boxed{-\frac{24}{7}}$

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  3. #3
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    thanks for explaining it clearly
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