Solveeit Logo
Login

Question

Question: Period of \(\sin\theta - \sqrt{3}\cos\theta\)is...

Period of sin⁡θ−3cos⁡θ\sin\theta - \sqrt{3}\cos\thetais

A

π4\frac{\pi}{4}

B

π2\frac{\pi}{2}

C

π\pi

D

2π2\pi

Answer

2π2\pi

Explanation

Solution

sin⁡θ−3cos⁡θ=2(12sin⁡θ−32.cos⁡θ)=2sin⁡(θ−π3)\sin\theta - \sqrt{3}\cos\theta = 2\left( \frac{1}{2}\sin\theta - \frac{\sqrt{3}}{2}.\cos\theta \right) = 2\sin\left( \theta - \frac{\pi}{3} \right)

Hence period =2π= 2\pi.