First slide

Multiple and sub- multiple Angles

Question

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

Moderate

A

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

$cosec α = cosec 2 α + cosec 4 α$
C

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

$8 \cos αcos 2 αcos 4 α = 1$

Solution

1. $\tan αtan 2 αtan 3 α = \tan 3 α - \tan 2 α - \tan α$
always holds good
2.
$\begin{array}{l} R.H.S. = \frac{\sin 4 α + \sin 2 α}{\sin 2 αsin 4 α} \\ = \frac{2 \sin 3 αcos α}{\sin 2 αsin 4 α} = \frac{2 \sin 4 α \cdot \cos α}{\sin 2 α \cdot \sin 4 α} \end{array}$
$= \frac{1}{\sin α} = cosec α (using π / 7 = α)$
Hence, (b) is correct.
Also,
3 . $\begin{matrix} \cos 2 α = \cos \frac{2 π}{7} = - \cos (π - \frac{5 π}{7}) \\ = - \cos (\frac{5 π}{7}) = - \cos 5 α \end{matrix}$
.
$\begin{matrix} \cos α + \cos 3 α + \cos 5 α = \frac{\sin 3 α}{\sin α} \cos 3 α \\ = \frac{\sin 6 α}{2 \sin α} = \frac{\sin \frac{6 π}{7}}{2 \sin \frac{π}{7}} \\ = \frac{\sin (π + \frac{π}{7})}{2 \sin \frac{π}{7}} = \frac{1}{2} \end{matrix}$
4.
$8 \cos αcos 2 αcos 4 α = \frac{\sin 8 α}{\sin α} = \frac{\sin \frac{8 π}{7}}{\sin \frac{π}{7}} = - \frac{\sin \frac{π}{7}}{\sin \frac{π}{7}} = - 1$

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