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.
317 exam-style questions on AQA A Level Maths 1.8 E: Trigonometry, covering 1.8.1 Trigonometric definitions, rules and radians, 1.8.2 Small angle approximations (A-level only), 1.8.3 Trigonometric functions and exact values, 1.8.4 Reciprocal and inverse trigonometric functions (A-level only), 1.8.5 Trigonometric identities, 1.8.6 Compound and double angle formulae (A-level only), 1.8.7 Solving trigonometric equations, 1.8.8 Proofs with trigonometric identities, and 1.8.9 Trigonometry in context. Each one has a worked solution and a mark scheme showing where the marks go.