Question

sin(θ) = − 9/41 where 3π/2 < θ < 2π how to find sin(2θ) sin(θ/2) cos(2θ) cos(θ/2) tan(2θ) tan(θ/2)

Answer 1

We know the angle 3π/2 < θ < 2π  that simply means that θ is the IV quadrant, where the sine is negative, and the cosine is positive,

now, we know the sine is -9/41.... which is the negative?  well, the hypotenuse, the denominator, is never negative because is just a radius unit, so the negative has to be the numerator of 9.

$\bf sin(\theta )=\cfrac{\stackrel{opposite}{-9}}{\stackrel{hypotenuse}{41}}\impliedby \textit{now let's find the \underline{adjacent side}} \\\\\\ \textit{using the pythagorean theorem}\\\\ c^2=a^2+b^2\implies \pm\sqrt{c^2-b^2}=a\qquad \begin{cases} c=hypotenuse\\ a=adjacent\\ b=opposite\\ \end{cases} \\\\\\ \pm\sqrt{41^2-(-9)^2}=a\implies \pm\sqrt{1600}=a \\\\\\ \pm 40=a\implies \stackrel{IV~quadrant}{+40=a}$

now, we know what the sine and cosine are for θ, let's use that then,

$\bf sin(2\theta)=2sin(\theta)cos(\theta)\\\\\\ sin(2\theta)=2\left( \frac{-9}{41} \right)\left( \frac{40}{41} \right)\implies sin(2\theta)=-\cfrac{720}{1681}\\\\ -------------------------------\\\\ cos\left(\cfrac{{{ \theta}}}{2}\right)=\pm \sqrt{\cfrac{1+cos({{ \theta}})}{2}} \\\\\\ cos\left(\cfrac{{{ \theta}}}{2}\right)=\pm \sqrt{\cfrac{1+\frac{40}{41}}{2}}\implies cos\left(\cfrac{{{ \theta}}}{2}\right)=\pm \sqrt{\cfrac{\quad \frac{81}{41}\quad }{2}}$

$\bf cos\left(\cfrac{{{ \theta}}}{2}\right)=\pm \sqrt{\cfrac{81}{82}}\implies cos\left(\cfrac{{{ \theta}}}{2}\right)=\pm \cfrac{\sqrt{81}}{\sqrt{82}}\implies cos\left(\cfrac{{{ \theta}}}{2}\right)=\pm \cfrac{9}{\sqrt{82}} \\\\\\ \textit{and rationalizing the denominator we'd get}\qquad cos\left(\cfrac{{{ \theta}}}{2}\right)=\pm \cfrac{9\sqrt{82}}{82}$

$\bf -------------------------------\\\\ tan\left(\cfrac{{{ \theta}}}{2}\right)= \begin{cases} \pm \sqrt{\cfrac{1-cos({{ \theta}})}{1+cos({{ \theta}})}} \\ \quad \\ \cfrac{sin({{ \theta}})}{1+cos({{ \theta}})} \\ \quad \\ \boxed{\cfrac{1-cos({{ \theta}})}{sin({{ \theta}})}} \end{cases}$

$\bf tan\left(\cfrac{{{ \theta}}}{2}\right)=\cfrac{1-\frac{40}{41}}{\frac{-9}{41}}\implies tan\left(\cfrac{{{ \theta}}}{2}\right)=\cfrac{\frac{1}{41}}{\frac{-9}{41}}\implies tan\left(\cfrac{{{ \theta}}}{2}\right)=\cfrac{1}{\underline{41}}\cdot \cfrac{\underline{41}}{-9} \\\\\\ tan\left(\cfrac{{{ \theta}}}{2}\right)=-\cfrac{1}{9}$