Sorry, I'm having trouble with this trigonometry question
$\endgroup$ 4Find $\sin(\theta/2)$, given that $\sin \theta = -4/5$ and $\theta$ terminates in $180^\circ<\theta<270^\circ$.
3 Answers
$\begingroup$$\sin ^2 \dfrac{\theta}{2} = \dfrac{1-\cos \theta}{2}$
If $\sin \theta = -4/5$ and the terminal side is in the third quadrant, draw a reference triangle in the third quadrant. Label the hypotenuse $5$ and the "opposite" side $-4$. You can easily solve for the adjacent side, as this is a "$3-4-5$" right triangle. Notice that the sign of the adjacent side is negative, by nature of being in the third quadrant.
Now the cosine is adjacent/hypotenuse $=-3/5$. Plug this into the half angle formula and take the positive root (why? Remember that sine is the y-coordinate of the point on the unit circle...).
$\endgroup$ 1 $\begingroup$Use $\cos x = \pm\sqrt{1- \sin^2} x$ to determine $\cos \theta$ and use $180^\circ<\theta<270^\circ$ to determine its sign.
If you divide $180^\circ<\theta<270^\circ$ by $2$, you get the range of $\theta/2$. Can you now use $\cos (2x) = 1 - 2\sin^2 (x)$ to determine $\sin (\theta/2)$?
$\endgroup$ $\begingroup$Here is an easy way: $\sin2\theta=2\sin{\theta}\cos{\theta}$. So $$\sin{\theta}=\sin\left(2\dfrac{{\theta}}{2}\right)=2\sin{\dfrac{\theta}{2}}\cos{\dfrac{\theta}{2}}=2\sin{\dfrac{\theta}{2}}\sqrt{1-\sin^2{\dfrac{\theta}{2}}}.$$
Since $\sin{\theta}=\dfrac{-4}{5}$, thus
$$2\sin{\dfrac{\theta}{2}}\sqrt{1-\sin^2{\dfrac{\theta}{2}}}=\dfrac{-4}{5},$$ which implies
$$4\sin^2{\dfrac{\theta}{2}}\left(1-\sin^2{\dfrac{\theta}{2}}\right)=\dfrac{16}{25}.$$
Can you finish it up? It's very easy, just solve for $\sin{\dfrac{\theta}{2}}$ which will give you $\pm\dfrac{2}{\sqrt5}$ or $\pm\dfrac{1}{\sqrt5}$. But since $180<\theta<270$ then $90<\dfrac{\theta}{2}<135$ and so $\sin{\dfrac{\theta}{2}}$ must be $\dfrac{2}{\sqrt5}$.
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