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Trigonometry and Modelling

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Question 8
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≤1800 \leq A \leq 1800≤A≤180, the equation,

3sin⁡A−4sin⁡3A=0.85 3\sin A - 4\sin^3 A = 0.85 3sinA−4sin3A=0.85

Give your answers to 1 decimal place.

[4]

Trigonometry and Modelling Questions

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