For α=π7 which of the following hold(s) good?

(A) 3α=2α+α tan3α=tan(2α+α) tan3α−tan2α−tanα=tanα tan2α tan3α B) RHS=sin4α+sin2αsin2α sin4α=2sin3α cosαsin2α sin4α=2sin3π7 cosπ7sin2π7sin4π7=1sinα=cosecα C) cosα−cos2α+cos3α=cosπ7−cos2π7+cos3π7=12 D) 8cosα cos2α cos4α=−1

For α=π7 which of the following hold(s) good?

(A) 3α=2α+α tan3α=tan(2α+α) tan3α−tan2α−tanα=tanα tan2α tan3α B) RHS=sin4α+sin2αsin2α sin4α=2sin3α cosαsin2α sin4α=2sin3π7 cosπ7sin2π7sin4π7=1sinα=cosecα C) cosα−cos2α+cos3α=cosπ7−cos2π7+cos3π7=12 D) 8cosα cos2α cos4α=−1

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For $α = \frac{π}{7}$ which of the following hold(s) good?

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$\cos α - \cos 2 α + \cos 3 α = \frac{1}{2}$

$8 \cos α \cos 2 α \cos 4 α = 1$

$\tan α \tan 2 α \tan 3 α = \tan 3 α - \tan 2 α - \tan α$

$\cos e c α = \cos e c 2 α + \cos e c 4 α$

answer is A, B, C.

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Detailed Solution

$(A) 3 α = 2 α + α \tan 3 α = \tan (2 α + α) \tan 3 α - \tan 2 α - \tan α = \tan α \tan 2 α \tan 3 α B) R H S = \frac{\sin 4 α + \sin 2 α}{\sin 2 α \sin 4 α} = \frac{2 \sin 3 α \cos α}{\sin 2 α \sin 4 α} = \frac{2 \sin \frac{3 π}{7} \cos \frac{π}{7}}{\sin \frac{2 π}{7} \sin \frac{4 π}{7}} = \frac{1}{\sin α} = \cos e c α C) \cos α - \cos 2 α + \cos 3 α = \cos \frac{π}{7} - \cos \frac{2 π}{7} + \cos \frac{3 π}{7} = \frac{1}{2} D) 8 \cos α \cos 2 α \cos 4 α = - 1$