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1.5 Trigonometry

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Question 90
a.

Solve the equation sin⁡2x=tan⁡2x\sin^2 x = \tan^2 xsin2x=tan2x for 0∘≤x≤180∘0^\circ \leq x \leq 180^\circ0∘≤x≤180∘

[5]
b.

Prove that 2sin⁡x−cos⁡2x+12+sin⁡x≡sin⁡x\displaystyle \frac{2 \sin x - \cos^2 x + 1}{2 + \sin x} \equiv \sin x2+sinx2sinx−cos2x+1​≡sinx

[3]

1.5 Trigonometry Questions

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