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Trigonometry and Modelling

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Question 6
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≤1800 \leq \theta \leq 1800≤θ≤180, the equation,

3sin⁡θ−4sin⁡3θ=0.4 3\sin\theta - 4\sin^3\theta = 0.4 3sinθ−4sin3θ=0.4

Give your answers to 1 decimal place.

[4]

Trigonometry and Modelling Questions

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