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Sorry, I'm having trouble with this trigonometry question

Find $\sin(\theta/2)$, given that $\sin \theta = -4/5$ and $\theta$ terminates in $180^\circ<\theta<270^\circ$.

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Do you know half angle formulas? It's just a direct application. – Euler....IS_ALIVE Apr 7 '13 at 20:51
That's a strange choice for a variable name. – TMM Apr 7 '13 at 20:52
I'm guessing it was intended to be $\theta$, which produced by using \theta. – Zev Chonoles Apr 7 '13 at 20:54

$\sin ^2 \dfrac{\theta}{2} = \dfrac{1-\cos \theta}{2}$

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...).

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So did you ever solve this? – The Chaz 2.0 Apr 16 '13 at 4:40

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 $\sin (\theta/2)$?

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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


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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sorry, is there a simpler way, cause im just lost. this looks nothing like I was taugh – flare Apr 7 '13 at 21:52
Show us how you start your working then. – YYG Apr 7 '13 at 21:55
I just don't know how to start. It's all confusing to me – flare Apr 7 '13 at 21:58
Okay can I know just what you have been treated under trigonometry so that I will work it that way? – YYG Apr 7 '13 at 22:04
We've gone over the sum and differences, double and half angle formulas, I just don't know how im suppose to use em in a question like this. I thought I could use the half angle formula sin x/2 = +_ squareroot by 1-cosx/2 but I don't have the cosine in the question. I do better when with charts – flare Apr 7 '13 at 22:08

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