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

or

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

using

cos A = 2cos^2 (A/2) -1

try the last one.

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