Thread: 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?

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 consider pythagorean triplet 5,12,13.

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?

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 ...



then ...

... quad I , correct?

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 -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.

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

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?
[tex] \sin\left(\frac{u}{2}\right)=\sqrt{\dfrac{1-\cos(u)}{2}} [/tex]
gives

yes i try to use latex.

i have given only the approach to the problem.