Here's the question you clicked on:
sunni0$U
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
|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?
I know them but I don't know how to use them.
\[\sin(\frac{x}{2})=\pm \sqrt{\frac{1}{2}(1-\cos(x))}\] \[\cos(\frac{x}{2})=\pm \sqrt{\frac{1}{2}(1+\cos(x))}\] \[\tan(\frac{x}{2})=\frac{\sin(\frac{x}{2})}{\cos(\frac{x}{2})}\]
I know how to get the decimal answers, but I need exact answers.
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
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?
\[\sin(\frac{x}{2})=-\sqrt{\frac{1}{2}(1-\frac{-12}{13})} =- \sqrt{\frac{1}{2}(1+\frac{12}{13})}=-\sqrt{\frac{1}{2}(\frac{13}{13}+\frac{12}{13})}\] \[=-\sqrt{\frac{1}{2}(\frac{25}{13})}=-\sqrt{\frac{25}{26}}=-\frac{\sqrt{25}}{\sqrt{26}}=-\frac{5}{\sqrt{26}}\] This answer can be simplified more by choosing to rationalize the denominator
Ok thanks. I figured out the tan x/2 part