## sunni0$U Group Title Find sin x/2, cos x/2, and tan x/2, if cos x = -12/13, 180 degrees is less than x which is less than 270 degrees one year ago one year ago • This Question is Closed 1. myininaya Group Title |dw:1352334915182:dw| But anyways, mainly I drew this to show what quadrant we are in. sine and cosine are both negative in this quadrant Do you know the half angle identities for sine and cosine? 2. sunni0$U Group Title

I know them but I don't know how to use them.

3. myininaya Group Title

$\sin(\frac{x}{2})=\pm \sqrt{\frac{1}{2}(1-\cos(x))}$ $\cos(\frac{x}{2})=\pm \sqrt{\frac{1}{2}(1+\cos(x))}$ $\tan(\frac{x}{2})=\frac{\sin(\frac{x}{2})}{\cos(\frac{x}{2})}$

4. sunni0$U Group Title I know how to get the decimal answers, but I need exact answers. 5. myininaya Group Title Since we are looking at the third quadrant and I said sine and cosine are both negative there, then we are actually looking at: $\sin(\frac{x}{2})=-\sqrt{\frac{1}{2}(1-\cos(x))}$ $\cos(\frac{x}{2})=-\sqrt{\frac{1}{2}(1+\cos(x))}$ Replace cos(x) with what it equals which is -12/13 then simplify 6. sunni0$U Group Title

For sin x/2 I got -sqrt of 25/26, which is wrong apparently, but I got cos x/2=-sqrt of 1/26 and it was right. Also, how do I find tan x/2 from all of this?

7. myininaya Group Title

$\sin(\frac{x}{2})=-\sqrt{\frac{1}{2}(1-\frac{-12}{13})} =- \sqrt{\frac{1}{2}(1+\frac{12}{13})}=-\sqrt{\frac{1}{2}(\frac{13}{13}+\frac{12}{13})}$ $=-\sqrt{\frac{1}{2}(\frac{25}{13})}=-\sqrt{\frac{25}{26}}=-\frac{\sqrt{25}}{\sqrt{26}}=-\frac{5}{\sqrt{26}}$ This answer can be simplified more by choosing to rationalize the denominator

8. sunni0\$U Group Title

Ok thanks. I figured out the tan x/2 part

9. myininaya Group Title

np.