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1.5.12 Double angle and compound angle formulae (A-level only)

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Question 3
a.

By writing sin⁡3θ \sin 3\theta\,sin3θ as sin⁡(2θ+θ)\sin(2\theta + \theta)sin(2θ+θ), show that sin⁡3θ=3sin⁡θ−4sin⁡3θ\sin 3\theta = 3\sin\theta - 4\sin^3\thetasin3θ=3sinθ−4sin3θ

[2]
b.

Solve, for 0≤θ≤π0 \leq \theta \leq \pi0≤θ≤π, the equation,

3sin⁡θ−4sin⁡3θ=0.5 3\sin\theta - 4\sin^3\theta = 0.5 3sinθ−4sin3θ=0.5

Give your answers in terms of π\piπ.

[4]

1.5.12 Double angle and compound angle formulae (A-level only) Questions

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