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Math Help - 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 \sin\left(\frac{x}{2}\right) = \frac{2}{3},\;\;0 \leq x \leq \frac{\pi}{2}

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

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


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

    Hence: . \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: . \sin\frac{\theta}{2} \;=\;\sqrt{\frac{1-\cos\theta}{2}}

    \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 \tan x = -\frac{3}{4},\;\;\frac{\pi}{2} \leq x \leq \pi

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

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


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

    (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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