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Quintuple-angle formula for sine used to find exact trigonometric values for multiples of pi over 5

Quintuple-angle formula for sine used to find exact trigonometric values for multiples of pi over 5

IB Year 6 | Grade 12
24 marks
  1. Use the expansion of \(\cos(A+B)\), together with suitable double-angle identities, to prove that \(\cos3\theta=4\cos^3\theta-3\cos\theta\).[3]
  2. It is given that \(\sin3\theta=3\sin\theta-4\sin^3\theta\). By writing \(5\theta=3\theta+2\theta\), prove that \(\sin5\theta=16\sin^5\theta-20\sin^3\theta+5\sin\theta\).[7]
  3. Use the equation \(\sin5\theta=0\) to solve \(16\sin^4\theta-20\sin^2\theta+5=0\) for \(0<\theta<\pi\), giving each answer as a rational multiple of \(\pi\).[4]
  4. Find the four exact values of \(\sin\theta\) that satisfy \(16\sin^4\theta-20\sin^2\theta+5=0\).[5]
    1. Hence show that \(\cos\dfrac{2\pi}{5}=\dfrac{\sqrt5-1}{4}\).[3]
    2. State the exact value of \(\sin\dfrac{3\pi}{5}\).[2]

Solution:

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Answer:(c) \(\theta=\dfrac{\pi}{5},\ \dfrac{2\pi}{5},\ \dfrac{3\pi}{5},\ \dfrac{4\pi}{5}\) (d) \(\sin\theta=\pm\sqrt{\dfrac{5-\sqrt5}{8}},\ \pm\sqrt{\dfrac{5+\sqrt5}{8}}\) (e)(ii) \(\sqrt{\dfrac{5+\sqrt5}{8}}\)

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