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3.4 Trigonometry (A-level only)

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Question 37
a.

By writing sin⁡3A \sin 3A\,sin3A as sin⁡(2A+A)\sin(2A + A)sin(2A+A), show that sin⁡3A=3sin⁡A−4sin⁡3A\sin 3A = 3\sin A - 4\sin^3 Asin3A=3sinA−4sin3A

[2]
b.

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

3sin⁡A−4sin⁡3A=12 3\sin A - 4\sin^3 A = \frac{1}{\sqrt{2}} 3sinA−4sin3A=2​1​

Give your answers in terms of π\piπ.

[4]

3.4 Trigonometry (A-level only) Questions

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