+3693 Prove tan(θ / 2) = sin θ / (1 + cos θ) for θ in quadrant 1 by filling in the reasons below.
+128485 tan (Θ/2) = sin Θ/ (1 + cosΘ)
sin(Θ/2)/cos(Θ/2) =
√ [(1-cosΘ)/2] / √[(1 + cosΘ)/2] =
√[(1 - cos Θ)/(1 + cos Θ)] = multiply top and bottom by √[(1 + cosΘ)/(1 + cosΘ)] =
√[(1 - cosΘ)(1+cosΘ)] / [1 + cosΘ] =
√[1 - cos^2 Θ] / (1 + cosΘ] =
√[sin^2 Θ] / (1 + cos Θ] =
sin Θ / [1 + cos Θ]
Notice, that since we are in the first quadrant, the tangent is positive everywhere except at 0 and pi/2....so the root wias positive.
+128485 tan (Θ/2) = sin Θ/ (1 + cosΘ)
sin(Θ/2)/cos(Θ/2) =
√ [(1-cosΘ)/2] / √[(1 + cosΘ)/2] =
√[(1 - cos Θ)/(1 + cos Θ)] = multiply top and bottom by √[(1 + cosΘ)/(1 + cosΘ)] =
√[(1 - cosΘ)(1+cosΘ)] / [1 + cosΘ] =
√[1 - cos^2 Θ] / (1 + cosΘ] =
√[sin^2 Θ] / (1 + cos Θ] =
sin Θ / [1 + cos Θ]
Notice, that since we are in the first quadrant, the tangent is positive everywhere except at 0 and pi/2....so the root wias positive.
