## he66666 5 years ago Trig Identity question? The angle x lies in the interval pi/2≤x≤pi, and sin²x=4/9. Determine cos(x/2). answer: √[(3-√5)/6] How do you solve this question (using trig identities)? It gets confusing because it's x/2, not just x. I keep getting 7/9 as my answer..

1. anonymous

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

2. anonymous

square root should be over whole thing.

3. he66666

How does it have a square root? :S Is it derived from a trig identity?

4. anonymous

$sin^2(x)=\frac{4}{9}$ $sin(x)=\pm \frac{3}{4}$

5. anonymous

derived from "double angle" formula $cos(2x)=2cos^2(x)-1$ replace 2x by x t so x gets replaced by $\frac{x}{2}$ and then solve for $cos(\frac{x}{2})$

6. anonymous

$cos(x)=2cos^2(\frac{x}{2})-1$ $cos(x)+1=2cos^2(\frac{x}{2})$ $\frac{cos(x)-1}{2}=cos^2(\frac{x}{2})$

7. anonymous

typo should be $\frac{cos(x)+1}{2}=cos^2(\frac{x}{2})$

8. anonymous

then take the square root of the whole thing. plus or minus of course. i cannot seem to type set it.

9. anonymous

in case $sin(x)=\frac{3}{4}$ do you know how to find $cos(x)$?

10. he66666

yes, pythagorean theorem?

11. anonymous

whoa hold the phone i am tired. it is $sin(x)=\frac{4}{9}$

12. anonymous

damn $sin^2(x)=\frac{4}{9}$ $sin(x)=\frac{2}{3}$

13. anonymous

draw a triangle opposite side 2 and hypotenuse 3 my mistake sorry

14. anonymous

adjacent side is $\sqrt{3^2-2^2}=\sqrt{9-4}=\sqrt{5}$

15. anonymous

so $cos(x)=\frac{\sqrt{5}}{3}$

16. anonymous

now plug in $\frac{\sqrt{5}}{3}$ in the formula up top to get your answer. i will write it if you like.

17. he66666

no it's alright, I got it. Thanks so much for the help satellite! :)

18. anonymous

oh another mistake i forgot x was in quad II so cosine is negative. it is $-\frac{\sqrt{5}}{3}$

19. anonymous

make sure you use $-\frac{\sqrt{5}}{3}$

20. he66666

alright, I will :) thanks

21. anonymous

welcome