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Vinil7 [7]
3 years ago
10

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)

Mathematics
1 answer:
jenyasd209 [6]3 years ago
5 0
We know the angle <span>3π/2 < θ < 2π  that simply means that </span>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}}&#10;\\\\\\&#10;\textit{using the pythagorean theorem}\\\\&#10;c^2=a^2+b^2\implies \pm\sqrt{c^2-b^2}=a\qquad &#10;\begin{cases}&#10;c=hypotenuse\\&#10;a=adjacent\\&#10;b=opposite\\&#10;\end{cases}&#10;\\\\\\&#10;\pm\sqrt{41^2-(-9)^2}=a\implies \pm\sqrt{1600}=a&#10;\\\\\\&#10;\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)\\\\\\&#10;sin(2\theta)=2\left( \frac{-9}{41} \right)\left( \frac{40}{41} \right)\implies sin(2\theta)=-\cfrac{720}{1681}\\\\&#10;-------------------------------\\\\&#10;cos\left(\cfrac{{{ \theta}}}{2}\right)=\pm \sqrt{\cfrac{1+cos({{ \theta}})}{2}}&#10;\\\\\\&#10;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}}&#10;\\\\\\&#10;

\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}}&#10;\\\\\\&#10;\textit{and rationalizing the denominator we'd get}\qquad cos\left(\cfrac{{{ \theta}}}{2}\right)=\pm \cfrac{9\sqrt{82}}{82}

\bf -------------------------------\\\\&#10;tan\left(\cfrac{{{ \theta}}}{2}\right)=&#10;\begin{cases}&#10;\pm \sqrt{\cfrac{1-cos({{ \theta}})}{1+cos({{ \theta}})}}&#10;\\ \quad \\&#10;\cfrac{sin({{ \theta}})}{1+cos({{ \theta}})}&#10;\\ \quad \\&#10;\boxed{\cfrac{1-cos({{ \theta}})}{sin({{ \theta}})}}&#10;\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}&#10;\\\\\\&#10;tan\left(\cfrac{{{ \theta}}}{2}\right)=-\cfrac{1}{9}
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