s i n θ c o s θ sin\theta cos\theta

Geometry Level 2

s i n θ + s i n 2 θ + s i n 3 θ = 1 + c o s θ + c o s 2 θ \large sin\theta+sin2\theta+sin3\theta=1+cos\theta+cos2\theta

Find the value of θ \theta in the range 0 < θ < π 2 0<\theta<\frac{\pi}{2}

π 3 \frac{\pi}{3} π 4 \frac{\pi}{4} π 2 \frac{\pi}{2} π 6 \frac{\pi}{6} There is no solution between this limit.

Md Mehedi Hasan by Md Mehedi Hasan , 22, Bangladesh

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1 solution

Md Mehedi Hasan
Nov 12, 2017

s i n θ + s i n 2 θ + s i n 3 θ = 1 + c o s θ + c o s 2 θ ⇒ 2 s i n 2 θ c o s θ + s i n 2 θ = c o s θ + 2 c o s 2 θ ⇒ s i n 2 θ ( 2 c o s θ + 1 ) − c o s θ ( 2 c o s θ + 1 ) = 0 ⇒ ( 2 c o s θ + 1 ) ( 2 s i n θ c o s θ − c o s θ ) = 0 ⇒ c o s θ ( 2 c o s θ + 1 ) ( 2 s i n θ − 1 ) = 0 sin\theta+sin2\theta+sin3\theta=1+cos\theta+cos2\theta\\ \Rightarrow 2sin2\theta cos\theta +sin2\theta=cos\theta+2cos^2\theta\\ \Rightarrow sin2\theta(2cos\theta+1)-cos\theta(2cos\theta+1)=0\\ \Rightarrow (2cos\theta+1)(2sin\theta cos\theta-cos\theta)=0\\ \Rightarrow cos\theta(2cos\theta+1)(2sin\theta-1)=0

For this interval, we have to take only

2 s i n θ − 1 = 0 ∴ θ = π 6 2sin\theta-1=0\\\therefore\theta=\boxed{\frac{\pi}{6}}

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