In this question you must show all stages of your working. Solutions relying entirely on calculator technology are not acceptable.
Two acoustic waves are superimposed such that their resulting phase angle ϕ\phiϕ satisfies the equation
6sin(ϕ−60∘)=2cos(ϕ+45∘) \sqrt{6} \sin(\phi - 60^\circ) = 2 \cos(\phi + 45^\circ) 6sin(ϕ−60∘)=2cos(ϕ+45∘)Show that
tanϕ=52+3 \tan \phi = \frac{5}{2 + \sqrt{3}} tanϕ=2+35and hence that
tanϕ=10−53 \tan \phi = 10 - 5\sqrt{3} tanϕ=10−53Hence or otherwise, solve for 0≤θ<180∘0 \le \theta < 180^\circ0≤θ<180∘,
6sin(3θ−60∘)=2cos(3θ+45∘) \sqrt{6} \sin(3\theta - 60^\circ) = 2 \cos(3\theta + 45^\circ) 6sin(3θ−60∘)=2cos(3θ+45∘)giving your answers to one decimal place.
27 exam-style questions on WJEC A Level Maths 3.5.6 Trigonometry (A-level only). Each one has a worked solution and a mark scheme showing where the marks go.