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Francesco Paucek

Feb 20, 2025

Tan theta=4/3 ….piNovember 17, 2021 by thanh

please help me find exact answer for cos(theta/2) and sin(theta/2)
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tan(θ) = 4/3 π < θ < 3π/2, θ is in QIII. To find cos(θ/2) and sin(θ/2), you only need to now where θ/2 is located (to know the signs of cos(θ/2) and sin(θ/2)) and you need to know cos(θ). Since θ/2 is in QIII, π < θ < 3π/2, π/2 < θ/2 < 3π/4 and θ/2 is in QII.cos(θ/2) = ±√((1 – cos(θ))/2) sin(θ/2) = ±√((1 + cos(θ))/2)cos(θ) = -3/5 since tan(θ) = 4/3 in QIII.Since θ/2 is in QII cos(θ/2) is negative and sin(θ/2) is positive.cos(θ/2) = -√((1 – -3/5)/2) sin(θ/2) = √((1 + -3/5))/2)cos(θ/2) = -√((1 + 3/5)/2) —–> cos(θ/2) = -√((8/5)/2) —–> cos(θ/2) = -√(8/10) —–> cos(θ/2) = -2/√5 —–> cos(θ/2) = -2√5/5sin(θ/2) = √((1 – 3/5))/2) —–> sin(θ/2) = √((2/5))/2 —–> sin(θ/2) = √(2/10) —–> sin(θ/2) = √(1/5) —-> sin(θ/2) = √5/5... Show More
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