Question

For 0 ≤ θ < 2 π what are the solutions to sin^2(θ) =2sin^2(θ/2)

Answer 1

Answer:

Option A: $\{0,\frac{\pi}{2},\frac{3\pi}{2}\}$

Step-by-step explanation:

$sin^2(\theta)=2sin^2(\frac{\theta}{2}), [0,2\pi)$

$sin^2(\theta)=2sin^2(\frac{\theta}{2})$

$sin^2(\theta)=2sin(\frac{\theta}{2})sin(\frac{\theta}{2})$

$sin^2(\theta)=2(\sqrt{\frac{1-cos(\theta)}{2}})(\sqrt{\frac{1-cos(\theta)}{2}})$

$sin^2(\theta)=2(\frac{1-cos(\theta)}{2})$

$sin^2(\theta)=\frac{2-2cos(\theta)}{2}$

$sin^2(\theta)=1-cos(\theta)$

$1-cos^2(\theta)=1-cos(\theta)$

$-cos^2(\theta)=-cos(\theta)$

$cos^2(\theta)=cos(\theta)$

$cos^2(\theta)-cos(\theta)=0$

$cos(\theta)[cos(\theta)-1]=0$

$cos(\theta)=0$

$\theta=\frac{\pi}{2},\frac{3\pi}{2}$

$cos(\theta)-1=0$

$cos(\theta)=1$

$\theta=0$

Therefore, the solutions contained within the interval are $\{0,\frac{\pi}{2},\frac{3\pi}{2}\}$

Helpful Tips:

Half-Angle Formula: $sin(\frac{\theta}{2})=\pm\sqrt{\frac{1-cos(\theta)}{2}}$

Pythagorean Identity: $sin^2(\theta)+cos^2(\theta)=1,sin^2(\theta)=1-cos^2(\theta),cos^2(\theta)=1-sin^2(\theta)$

Answer 2

$\sin^2 \theta = 2 \sin^2 \left(\dfrac{\theta}2 \right)\\\\\implies \sin^2 \theta = 1- \cos 2 \cdot \dfrac{\theta}2\\\\\implies \sin^2 \theta = 1- \cos \theta \\\\\implies 1-\cos^2 \theta = 1 - \cos \theta \\\\\implies -\cos^2 \theta - \cos \theta = 0\\\\\implies \cos^2 \theta - \cos \theta = 0\\\\\implies \cos \theta( \cos \theta -1) = 0\\\\\\\text{Now,}\\\\\cos \theta = 0\\\\\implies \theta = n\pi + \dfrac{\pi}2\\\\\text{For n = 0,1 and}~ [0.2\pi)$

$\theta = \dfrac{\pi}2, \dfrac{3\pi}2$

$\text{Again,} \\\\\cos \theta -1= 0\\\\\implies \cos \theta = 1\\\\\implies \theta = 2n\pi\\\\\text{For n= 0 and}~ [0,2\pi)\\\\\theta = 0\\\\\text{Combine solutions,}\\\\\theta = 0, \dfrac{\pi}2, \dfrac{3\pi}2$