Name: Trigonometric Identities
Uploaded: 2019-07-19
Duration: 7 min 30 s
Description: Just like all superheroes have their own identities, trigonometric ratios too have their own identities! Don't you want to know what they are? Let's find out.

Question

If

A

is not an integral multiple of

π

, prove that

\cos A \cdot \cos 2 A \cdot \cos 4 A \cdot \cos 8 A = \frac{\sin 16 A}{16 \sin A}

and hence deduce that

\cos \frac{2 π}{15} \cdot \cos \frac{4 π}{15} \cdot \cos \frac{8 π}{15} \cdot \cos \frac{16 π}{15} = \frac{1}{16}

.

Accepted Answer

To Prove: $\cos A \cdot \cos 2 A \cdot \cos 4 A \cdot \cos 8 A = \frac{\sin 16 A}{16 \sin A}$ , where $A$ is not an integral multiple of $π$

L.H.S. $= \cos A \cdot \cos 2 A \cdot \cos 4 A \cdot \cos 8 A$

$= \frac{1}{2 \sin A} (2 \sin A \cos A) \cdot \cos 2 A \cdot \cos 4 A \cdot \cos 8 A$ [Multiply and divide by $2 \sin A$ ]

$= \frac{1}{2 \sin A} \sin 2 A \cdot \cos 2 A \cdot \cos 4 A \cdot \cos 8 A$ [Since $\sin 2 θ = 2 \sin θ \cos θ$ ]

$= \frac{1}{2^{2} \sin A} (2 \sin 2 A \cos 2 A) \cdot \cos 4 A \cdot \cos 8 A$

$= \frac{1}{2^{2} \sin A} (\sin 4 A) \cdot \cos 4 A \cdot \cos 8 A$

$= \frac{1}{2^{3} \sin A} (2 \sin 4 A \cos 4 A) \cdot \cos 8 A$

$= \frac{1}{2^{3} \sin A} \sin 8 A \cos 8 A$

$= \frac{1}{2^{4} \sin A} 2 \sin 8 A \cos 8 A$

$= \frac{1}{16 \sin A} \sin 16 A$

Hence, $\cos A \cdot \cos 2 A \cdot \cos 4 A \cdot \cos 8 A = \frac{\sin 16 A}{16 \sin A} \to (1)$

Put $A = \frac{2 π}{15}$ in $(1)$ , we get

$\cos \frac{2 π}{15} \cdot \cos 2 (\frac{2 π}{15}) \cdot \cos 4 (\frac{2 π}{15}) \cdot \cos 8 (\frac{2 π}{15}) = \frac{\sin 16 (\frac{2 π}{15})}{16 \sin (\frac{2 π}{15})}$

$\Rightarrow \cos \frac{2 π}{15} \cdot \cos (\frac{4 π}{15}) \cdot \cos (\frac{8 π}{15}) \cdot \cos (\frac{16 π}{15}) = \frac{\sin \frac{32 π}{15}}{16 \sin \frac{2 π}{15}}$

$= \frac{\sin (2 π + \frac{2 π}{15})}{16 \sin \frac{2 π}{15}}$

$= \frac{\sin \frac{2 π}{15}}{16 \sin \frac{2 π}{15}}$

$= \frac{1}{16}$

Hence, $\cos \frac{2 π}{15} \cdot \cos \frac{4 π}{15} \cdot \cos \frac{8 π}{15} \cdot \cos \frac{16 π}{15} = \frac{1}{16}$ .

If $A$ is not an integral multiple of $π$ , prove that $\cos A \cdot \cos 2 A \cdot \cos 4 A \cdot \cos 8 A = \frac{\sin 16 A}{16 \sin A}$ and hence deduce that $\cos \frac{2 π}{15} \cdot \cos \frac{4 π}{15} \cdot \cos \frac{8 π}{15} \cdot \cos \frac{16 π}{15} = \frac{1}{16}$ .

Important Questions on Trigonometric Ratios upto Transformations

Learn and Prepare for any exam you want !

If A is not an integral multiple of π, prove that cosA·cos2A·cos4A·cos8A=sin16A16sinA and hence deduce that cos2π15·cos4π15·cos8π15·cos16π15=116.

Important Questions on Trigonometric Ratios upto Transformations

Learn and Prepare for any exam you want !

If $A$ is not an integral multiple of $π$ , prove that $\cos A \cdot \cos 2 A \cdot \cos 4 A \cdot \cos 8 A = \frac{\sin 16 A}{16 \sin A}$ and hence deduce that $\cos \frac{2 π}{15} \cdot \cos \frac{4 π}{15} \cdot \cos \frac{8 π}{15} \cdot \cos \frac{16 π}{15} = \frac{1}{16}$ .