SOLUTION: Find the exact values of sin(θ/2), cos(θ/2), and tan(θ/2) for the given conditions. sec θ = −4; 180° < θ < 270° I found sin(θ/2) to be {{{sqrt(10)}}}/2. For cos(θ

Algebra ->  Trigonometry-basics -> SOLUTION: Find the exact values of sin(θ/2), cos(θ/2), and tan(θ/2) for the given conditions. sec θ = −4; 180° < θ < 270° I found sin(θ/2) to be {{{sqrt(10)}}}/2. For cos(θ      Log On


   

Question 1159667: Find the exact values of sin(θ/2), cos(θ/2), and tan(θ/2) for the given conditions.
sec θ = −4; 180° < θ < 270°
I found sin(θ/2) to be sqrt%2810%29/2. For cos(θ/2) I got sqrt%286%29/4 which is wrong and for tan(θ/2) I got sqrt%2815%29/3 which is wrong too. What did I do wrong???
This is my work:
cos(θ/2)= -%28sqrt%28%281%2B%28-1%2F4%29%29%2F2%29%29= sqrt%283%2F8%29= sqrt%283%29/2sqrt%282%29= sqrt%286%29/4
tan(θ/2)=sqrt%28%28%281%2B1%2F4%29%2F%281-1%2F4%29%29%29= sqrt%285%29/sqrt%283%29= sqrt%2815%29/3
Answer by solver91311(24713) About Me  (Show Source):
You can put this solution on YOUR website!


Half of an angle in Quadrant 3 must, perforce be located in Quadrant 2. If then . Furthermore, If then .

The cosine in QII is negative and the sine is positive in QII. Therefore tangent is also negative in QII. You had the negative sign in your cosine calculation, but you dropped it for some reason. And you didn't put it into your tangent calculation at all.

The devil, they say, is in the details.


John

My calculator said it, I believe it, that settles it