Double- and Half-Angle IdentitiesDate: 11/24/98 at 23:34:03 From: Jeff Hefner Subject: Double- and Half-Angle Identities I am having trouble understanding the purpose of double and half angles. For example, use double-angle or half-angle identities to find the exact value of each: 1) sin 5pi/12 2) cos 7pi/8 Date: 11/25/98 at 11:58:17 From: Doctor Anthony Subject: Re: Double- and Half-Angle Identities In general, we use the double- and half-angle formulas when we want to calculate the sine, cosine, etc, of an angle whose value isn't readily known. For example, we know that cos(30) = sqrt(3)/2. But what is cos(15)? Recall the half-angle formula for cosine: cos(2 theta) = cos^2(theta) - sin^2(theta) = 2 cos^2(theta) - 1 For sine it is: sin(2 theta) = 2 sin(theta) cos(theta) Then we know that sqrt(3)/2 = cos(30) = 2 cos^2(15) - 1. So: cos(15) = sqrt((1 + sqrt(3))/4) Turning to your first problem, for the first one, note in degrees sin(5 pi/12) = sin(75) = cos(15). Then we have: cos(30) = 2 cos^2(15) - 1 sqrt(3)/2 = 2 cos^2(15) - 1 2 cos^2(15) = 1 + sqrt(3)/2 2+sqrt(3) cos^2(15) = ---------- 4 sqrt(2+sqrt(3)) cos(15) = --------------- 2 So we must find sqrt(2+sqrt(3)). Consider: [sqrt(3)+1]^2 = 3 + 1 + 2 sqrt(3) = 4 + 2 sqrt(3) = 2[2 + sqrt(3)] and so: 2 + sqrt(3) = [sqrt(3) + 1]^2/2 sqrt([2+sqrt(3)]) = [sqrt(3)+1]/sqrt(2) Therefore: sqrt(3) + 1 sin (5 pi/12) = cos(15) = ----------- 2 sqrt(2) For the second problem, cos(7pi/8): In degrees cos(7.pi/8) = -cos(22.5). Thus: 2 tan(22.5) tan(45) = ---------------- = 1 1 - tan^2(22.5) 2 tan(22.5) = 1 - tan^2(22.5) tan^2(22.5) + 2 tan(22.5) - 1 = 0 tan(22.5) = [-2 + sqrt(4+4)]/2 = [-2+ 2sqrt(2)]/2 = sqrt(2)-1 tan^2(22.5) = 2 + 1 - 2 sqrt(2) = 3 - 2 sqrt(2) sec^2(22.5) = 1 + tan^2(22.5) = 4 - 2 sqrt(2) cos^2(22.5) = 1/[4 - 2 sqrt(2)] So cos(7 pi/8) = -1/[sqrt(4 - 2 sqrt(2)] - Doctor Anthony, The Math Forum http://mathforum.org/dr.math/ |
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