Find all solutions in the interval [0,2π)

label Calculus
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tan(a/2) = sin(a)

Mar 17th, 2015

tan(a/2) = +/-sqrt[(1-cosa)/1+cosa) = sin(a)

Square both sides to get

(1 - cosa)/(1+cosa) = sin^2a    but sin^2a = 1 - cos^2a

So,  (1 - cosa)/(1+cosa) = 1 - cos^2 = (1 + cosa)(1 - cosa)

therefore,

(1+cosa)^2 = 1

1+cosa = 1

cosa = 0

a = pi/2, 3pi/2

Mar 17th, 2015

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