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Trigonometric Identities and Equations

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Question 106
a.

Show that 2cos⁡2θ+5sin⁡θ−42−sin⁡θ≡2sin⁡θ−1\displaystyle \frac{2\cos^2 \theta + 5\sin \theta - 4}{2 - \sin \theta} \equiv 2\sin \theta - 12−sinθ2cos2θ+5sinθ−4​≡2sinθ−1

[4]
b.

Hence solve, for 0≤θ<360∘0 \leq \theta < 360^\circ0≤θ<360∘, the equation, 2cos⁡2θ+5sin⁡θ−42−sin⁡θ=2cos⁡θ−1\displaystyle \frac{2\cos^2 \theta + 5\sin \theta - 4}{2 - \sin \theta} = 2\cos \theta - 12−sinθ2cos2θ+5sinθ−4​=2cosθ−1

[3]
Markscheme

Trigonometric Identities and Equations Questions

  1. A Level
  2. /Maths
  3. /Trigonometric Identities and Equations

322 exam-style questions on Edexcel A Level Maths Trigonometric Identities and Equations, covering 10.1 Angles in all four Quadrants, 10.2 Exact Values of Trigonometric Ratios, 10.3 Trigonometric Identities, 10.4 Solving Trigonometric Equations, 10.5 Harder Trigonometric Equations, and 10.6 Equations and Identities. Each one has a worked solution and a mark scheme showing where the marks go.

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