# Math Help - Need help with 2 Trigonometric Questions

1. ## 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

2. 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: . $\sin x = \frac{3}{5},\;\cos x = -\frac{4}{5}$

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

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

3. thanks for explaining it clearly