Question

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..

Answer 1

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

Answer 2

square root should be over whole thing.

Answer 3

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

Answer 4

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

Answer 5

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
\[cos(\frac{x}{2})\]

Answer 6

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

Answer 7

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

Answer 8

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

Answer 9

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

Answer 10

yes, pythagorean theorem?

Answer 11

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

Answer 12

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

Answer 13

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

Answer 14

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

Answer 15

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

Answer 16

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

Answer 17

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

Answer 18

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

Answer 19

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

Answer 20

alright, I will :) thanks

Answer 21

welcome