Mark M. answered • 01/14/17
Tutor
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(955)
Retired Math prof with teaching and tutoring experience in trig.
sinx = 24/25
cos2x + sin2x = 1 So, cosx = ±√(1-sin2x) = ±7/25
sin(x/2) = ±√[(1-cosx)/2]
= ±√[(1-7/25)/2] or ±√[(1+7/25)/2]
= ±3/5 or ±4/5
Mark M.
tutor
The problem as stated does not restrict x to the interval [0, π]. As far as I can tell, x can be any real number for which sinx = 24/25.
Converting to degree measure:
Sin-1(24/25) ≈ 73.7°
sin(73.7°+360°) = sin(433.7°) = 24/25 also
If we let x = 433.7° then x/2 = 216.85°, which lies in Quadrant 3.
So, in this case, sinx > 0 but sin(x/2) < 0.
Mark M (Bayport, NY)
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01/14/17
James B.
I see all but one thing. Why " or +/- sq root [ (1+7/25)/2]" . Isn't thiis the half angle formula for the cos x/2?
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01/15/17
Mark M.
tutor
sin(x/2) = ±√[(1-cosx)/2]. cosx = ±7/25.
So, sin(x/2) = ±√[(1-7/25)/2] OR ±√[(1-(-7/25))/2]
= ±√[(1-7/25)/2] or ±√[(1+7/25)/2] = ±3/5 or ±4/5
Mark M (Bayport, NY)
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01/15/17
James B.
I see. Thank you so much!
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01/15/17
James B.
Thank you!
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01/15/17
Mark M.
0 < x/2 < π/2
x/2 is in QI and therefore sin x is positive.
01/14/17