Solve 2sin2θ = 1 + cos2θ for 0° ≤ θ ≤ 180°

sin2θ = 2sinθcosθ (double angle formula for sine)cos2θ = cos2θ - sin2θ (double angle formula for cosine) = 2cos2θ - 1 (utilising the trignometric identity sin2θ + cos2θ = 1)
We substitute these into our original equation to get4sinθcosθ = 1 + 2cos2θ - 1
Rearranging we getcosθ(cosθ - 2sinθ) = 0
Therefore, either cosθ = 0 or cosθ - 2sinθ = 0.
cosθ = 0 when θ = 90° only in the range we are given (draw a diagram)
We can divide the right hand equation by cosθ (note this is only valid when cosθ ≠ 0 <--> when θ ≠ 90°)This gives us1 - 2tanθ = 0 (Use inequality tanθ = sinθ/cosθ)
Which rearrages totanθ = 1/2Soθ = 26.6° only (draw diagram)(Note θ ≠ 90° so this was all valid)

We have θ = 26.6° and 90° as solutions.

Answered by Samuel N. Maths tutor

11854 Views

See similar Maths A Level tutors

Related Maths A Level answers

All answers ▸

Given that z=sin(x)/cos(x), show that dz/dx = sec^2(x).


i) Simplify (2 * sqrt(7))^2 ii) Find the value of ((2 * sqrt(7))^2 + 8)/(3 + sqrt(7)) in the form m + n * sqrt(7) where n and m are integers.


What is integration?


How do you differentiate y=cox(x)/sin(x)?


We're here to help

contact us iconContact usWhatsapp logoMessage us on Whatsapptelephone icon+44 (0) 203 773 6020
Facebook logoInstagram logoLinkedIn logo

© MyTutorWeb Ltd 2013–2025

Terms & Conditions|Privacy Policy
Cookie Preferences