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Sum of Trigonometric Ratios in Terms of Their Product

For α =π7 w...

Question

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

A

tan α tan 2 α tan 3 α = tan 3 α - tan 2 α - tan α

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B

cosec α = cosec 2 α + cosec 4 α

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C

cos α - cos 2 α + cos 3 α = 1 / 2

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D

8 cos α cos 2 α cos 4 α = 1

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Solution

The correct options are
A $tan α tan 2 α tan 3 α = tan 3 α - tan 2 α - tan α$
B $cosec α = cosec 2 α + cosec 4 α$
D $cos α - cos 2 α + cos 3 α = 1 / 2$
$α + 2 α = 3 α tan (α + 2 α) = tan (3 α) \frac{tan α + tan 2 α}{1 - tan α tan 2 α} = tan 3 α tan α + tan 2 α = tan 3 α - tan α tan 2 α tan 3 α tan α tan 2 α tan 3 α = tan 3 α - tan α - tan 2 α$

$cosec 2 α + cosec 4 α = \frac{1}{sin 2 α} + \frac{1}{sin 4 α} = \frac{sin 4 α + sin 2 α}{sin 2 α sin 4 α} = \frac{2 sin 3 α cos α}{2 sin α cos α sin 4 α} = \frac{sin 3 α}{sin α sin 4 α} = \frac{sin \frac{3 π}{7}}{sin \frac{π}{7} sin \frac{4 π}{7}} = \frac{sin (π - \frac{4 π}{7})}{sin \frac{π}{7} sin \frac{4 π}{7}} = \frac{sin \frac{4 π}{7}}{sin \frac{π}{7} sin \frac{4 π}{7}} = \frac{1}{sin \frac{π}{7}} = cosec \frac{π}{7} = cosec α$

$cos α - cos 2 α + cos 3 α = cos α + cos 3 α - cos 2 α = 2 cos 2 α cos α - cos 2 α = cos 2 α (2 cos α - 1) = cos 2 α (2 cos α - 1) . \frac{sin α}{sin α} = \frac{cos 2 α (2 sin α cos α - sin α)}{sin α} = \frac{cos 2 α (sin 2 α - sin α)}{sin α} = \frac{cos 2 α (2 cos \frac{3 α}{2} sin \frac{α}{2})}{2 sin \frac{α}{2} cos \frac{α}{2}} = \frac{cos 2 α cos \frac{3 α}{2}}{cos \frac{α}{2}} = \frac{1}{2} ⎛ ⎜ ⎜ ⎜ ⎝ \frac{2 cos \frac{2 π}{7} cos \frac{3 π}{14}}{cos \frac{π}{14}} ⎞ ⎟ ⎟ ⎟ ⎠ = \frac{1}{2} ⎛ ⎜ ⎜ ⎜ ⎝ \frac{2 cos (\frac{π}{2} - \frac{3 π}{14}) cos \frac{3 π}{14}}{cos \frac{π}{14}} ⎞ ⎟ ⎟ ⎟ ⎠ = \frac{1}{2} ⎛ ⎜ ⎜ ⎜ ⎝ \frac{2 sin \frac{3 π}{14} cos \frac{3 π}{14}}{cos \frac{π}{14}} ⎞ ⎟ ⎟ ⎟ ⎠ = \frac{1}{2} \frac{sin \frac{6 π}{14}}{cos \frac{π}{14}} = \frac{1}{2} \frac{sin (\frac{π}{2} - \frac{π}{14})}{cos \frac{π}{14}} = \frac{1}{2} \frac{cos \frac{π}{14}}{cos \frac{π}{14}} = \frac{1}{2}$

$8 cos α cos 2 α cos 4 α = \frac{4 (2 sin α cos α) cos 2 α cos 4 α}{sin α} = \frac{2 (2 sin 2 α cos 2 α) cos 4 α}{sin α} = \frac{2 sin 4 α cos 4 α}{sin α} = \frac{sin 8 α}{sin α} = \frac{sin \frac{8 π}{7}}{sin \frac{π}{7}} = \frac{sin (π + \frac{π}{7})}{sin \frac{π}{7}} = \frac{- sin \frac{π}{7}}{sin \frac{π}{7}} = - 1$

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