Posted by Bob on Friday, November 11, 2011 at 11:32am.
If A is between 270degrees and 360degrees, and cos A=12/13, find the exact value of:
sin2a, cos2a, sin1/2a, cos1/2a
trig/math - Steve, Friday, November 11, 2011 at 12:18pm
since acos(12/13) = .39479, a = 2*pi - .39479 = 5.88839
2a = 11.77678 is in the 4th quadrant, so
sin 2a = 2sin(a)cos(a) = 2(-5/13)(12/13) = -120/169
cos(2a) = cos^2(a) - sin^2(a) = 144/169 - 25/169 = 119/169
a/2 is in the 2nd quadrant
sin(a/2) = √(1/26)
cos(a/2) = -√(25/26)
trig/math - Reiny, Friday, November 11, 2011 at 12:20pm
if cosA = 12/13 in quadr IV, then
sinA = -5/13
sin 2A = 2sinAcosA = 2(12/13)(-5/13) = -120/169 =
cos 2A = cos^2 A - sin^2 A
= 144/169 - 25/169 = 119/169
we know cos2A = 1 - 2sin^2 A
cos A = 1 - 2sin^2 (A/2)
12/13 = 1 - 2sin^2 (A/2)
2sin^2 (A/2) = 1 - 12/13 = 1/13
sin^2 (A/2) = 1/26
sin (A/2) = ±1/√26
but if A is in IV, then A/2 is in II
so sin(A/2) = 1/√26
cos A = 2cos^2 (A/2) -1
try the last one.
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