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Trigonometry and Modelling

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Question 12
a.

Show that the equation [6sin⁡θcos⁡θcos⁡θ+sin⁡θ=(4+2sec⁡2θ)(cos⁡θ−sin⁡θ)[\frac{6 \sin \theta \cos \theta}{\cos \theta + \sin \theta} = (4 + 2\sec 2\theta)(\cos \theta - \sin \theta)[cosθ+sinθ6sinθcosθ​=(4+2sec2θ)(cosθ−sinθ) can be written in the form 3sin⁡2θ−4cos⁡2θ=23 \sin 2\theta - 4 \cos 2\theta = 23sin2θ−4cos2θ=2

[5]
b.

Hence solve for 0<x<π0 < x < \pi0<x<π [6sin⁡xcos⁡xcos⁡x+sin⁡x=(4+2sec⁡2x)(cos⁡x−sin⁡x)[\frac{6 \sin x \cos x}{\cos x + \sin x} = (4 + 2\sec 2x)(\cos x - \sin x)[cosx+sinx6sinxcosx​=(4+2sec2x)(cosx−sinx) giving your answers to 3 significant figures.

[5]

Trigonometry and Modelling Questions

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