Asked by Nikki
Use the half-angle identities to find all solutions on the interval [0,2pi) for the equation
cos^2(x) = sin^2(x/2)
cos^2(x) = sin^2(x/2)
Answers
Answered by
Damon
sin (x/2 ) = + sqrt([1-cos x]/2) if x/2 in q 1 or 2 thus if 0 < x < 2 pi
sin^2 (x/2) = [1-cos x]/2
so
cos^2 x = [1 - cos x]/2
2 cos^2 x = 1 - cos x
2 cos^2 x + cos x = 1
let z = cos x
2 z^2 + z -1 = 0
(2 z - 1)(z +1) = 0
z = cos x = 1/2
or cos x = -1
x = 60 deg (pi/3) or -60 deg (5 pi/3)
or x = 180 deg (pi)
sin^2 (x/2) = [1-cos x]/2
so
cos^2 x = [1 - cos x]/2
2 cos^2 x = 1 - cos x
2 cos^2 x + cos x = 1
let z = cos x
2 z^2 + z -1 = 0
(2 z - 1)(z +1) = 0
z = cos x = 1/2
or cos x = -1
x = 60 deg (pi/3) or -60 deg (5 pi/3)
or x = 180 deg (pi)
There are no AI answers yet. The ability to request AI answers is coming soon!
Submit Your Answer
We prioritize human answers over AI answers.
If you are human, and you can answer this question, please submit your answer.