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Cosine double angle identity in terms of tangent, a trigonometric equation and the exact value of tan 15 degrees

Cosine double angle identity in terms of tangent, a trigonometric equation and the exact value of tan 15 degrees

15 marks
  1. Show that \(\dfrac{1-\tan^2\theta}{1+\tan^2\theta}=\cos2\theta\) for \(\cos\theta\neq0\).[3]
  2. Hence, for \(0^\circ\leq\theta\leq360^\circ\), solve \(\dfrac{1-\tan^2\theta}{1+\tan^2\theta}=-\dfrac12\).[5]

It is given that \(t=\tan15^\circ\).

  1. By using part (a) with a suitable value of \(\theta\), show that \(t^2=7-4\sqrt3\).[4]
  2. Hence, find \(\tan15^\circ\) exactly, in the form \(p+q\sqrt3\) where \(p\) and \(q\) are integers.[3]

Solution:

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Answer:(a) \(\dfrac{\cos^2\theta-\sin^2\theta}{\cos^2\theta+\sin^2\theta}=\cos2\theta\) (b) \(\theta=60^\circ,\ 120^\circ,\ 240^\circ,\ 300^\circ\) (c) \(t^2=\dfrac{2-\sqrt3}{2+\sqrt3}=7-4\sqrt3\) (d) \(2-\sqrt3\)

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