# Thread: Location of half angle

1. ## Location of half angle

Sin u = 5/13 and pi/2<u<pi.

Find Sin (u/2); How do you determine the sign (positive or negative) for the value?

Using that information find cos (u/2): I have sqrt(26)/26. Is this positive or negative? How do I know?

2. Originally Posted by IDontunderstand
Sin u = 5/13 and pi/2<u<pi.

Find Sin (u/2); How do you determine the sign (positive or negative) for the value?
Which quadrant will it be in?

for $\cos u$ consider pythagorean triplet 5,12,13.

3. Originally Posted by IDontunderstand
Sin u = 5/13 and pi/2<u<pi.

Find Sin (u/2); How do you determine the sign (positive or negative) for the value?

Using that information find cos (u/2): I have sqrt(26)/26. Is this positive or negative? How do I know?
$\cos\left(u\right)=\frac{-12}{13}$

$\sin\left(\frac{u}{2}\right)=\sqrt{\frac{1-\cos(u)}{2}}$

4. Originally Posted by IDontunderstand
Sin u = 5/13 and pi/2<u<pi.

Find Sin (u/2); How do you determine the sign (positive or negative) for the value?

Using that information find cos (u/2): I have sqrt(26)/26. Is this positive or negative? How do I know?
if ...

$\displaystyle \frac{\pi}{2} < u < \pi$

then ...

$\displaystyle \frac{\pi}{4} < \frac{u}{2} < \frac{\pi}{2}$ ... quad I , correct?

5. sin u =5/13 and pi/2<u<pi

so find cos u = -12/13 as it is in 2nd quadrant.

use cos u= 2 $cos^2 u/2$ -1 to evaluate cos u/2.

you take only positive value as it is in 1st quadrant.

using that you can find sin u/2 also.

since u/2(pi/4<u/2<pi/2 ) is in 1st quadrant

both cos u/2 and sin u/2 are positive.

6. @saravananbs
Why not learn to post in symbols? You can use LaTeX tags
You do understand that this is not a homework service?
$$\sin\left(\frac{u}{2}\right)=\sqrt{\dfrac{1-\cos(u)}{2}}$$
gives $\sin\left(\frac{u}{2}\right)=\sqrt{\dfrac{1-\cos(u)}{2}}$

7. Originally Posted by Plato
@saravananbs
Why not learn to post in symbols? You can use LaTeX tags
You do understand that this is not a homework service?
$$\sin\left(\frac{u}{2}\right)=\sqrt{\dfrac{1-\cos(u)}{2}}$$
gives $\sin\left(\frac{u}{2}\right)=\sqrt{\dfrac{1-\cos(u)}{2}}$
yes i try to use latex.

i have given only the approach to the problem.