Further Trig Questions (1 Viewer)

[bujolover]

Hi guys,
1) Prove that ===>

I tried doing the technique where you solve the LHS and RHS separately and try and get them to one common value, but I got two very different values when doing this.

2) Solve the following equation for 0≤θ≤2π: 2cos2θ = tan(5π/4)

I got π/6, 5π/6, 7π/6 or 11π/6, but the given answer excludes the last two solutions (no working given). Am I wrong or is the given answer wrong?

Thanks in advance!

 

[He-Mann]

1) Tried to prove, didn't seem right. Plugged in A = 40, B = 5 degrees and realised the identity was wrong. It's meant to be



2) You are correct. Their answers are correct under the assumption that 0≤θ≤π