Question

# Prove the following identity:$\sin 6{}^\circ \sin 42{}^\circ \sin 66{}^\circ \sin 78{}^\circ =\dfrac{1}{16}$

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Hint: Convert all ‘sin’ into ‘cos’ except $\sin 6{}^\circ$, then make an adjustment so that you can use the formula $\sin 2x=2\times \sin x\times \cos x$ every time till we don’t get the answer.

To prove the equation given above we will write the equation first,
$\sin 6{}^\circ \sin 42{}^\circ \sin 66{}^\circ \sin 78{}^\circ =\dfrac{1}{16}$
Consider the Left Hand Side of the equation,
L.H.S. (Left Hand Side) $=\sin 6{}^\circ \sin 42{}^\circ \sin 66{}^\circ \sin 78{}^\circ$
As we all know that any angle $\theta$ can be replaced by $\left[ 90-\left( 90-\theta \right) \right]$ therefore we can write L.H.S. as shown below,
L.H.S. (Left Hand Side) $=\sin 6{}^\circ \times \sin \left[ 90{}^\circ -\left( 90{}^\circ -42{}^\circ \right) \right]\sin \left[ 90{}^\circ -\left( 90{}^\circ -66{}^\circ \right) \right]\times \sin \left[ 90{}^\circ -\left( 90{}^\circ -78{}^\circ \right) \right]$
Therefore by simplification we can write,
L.H.S. (Left Hand Side) $=\sin 6{}^\circ \times \sin \left[ 90{}^\circ -48{}^\circ \right]\times \sin \left[ 90{}^\circ -24{}^\circ \right]\times \sin \left[ 90{}^\circ -12{}^\circ \right]$
To proceed further in the solution we should know the formula given below,
Formula:
$\sin \left( 90{}^\circ -\theta {}^\circ \right)=\cos \theta {}^\circ$
By using the above formula we can write the L.H.S. as shown below,
$\therefore$ L.H.S. (Left Hand Side) $=\sin 6{}^\circ \times \cos 48{}^\circ \times \cos 24{}^\circ \times \cos 12{}^\circ$
On rearranging the above equation we will get,
$\therefore$ L.H.S. (Left Hand Side) $=\sin 6{}^\circ \times \cos 12{}^\circ \times \cos 24{}^\circ \times \cos 48{}^\circ$
If we observe the above equation carefully then we can say that the equation can be arranged such that it will take the form of sin2x.
Therefore we will multiply and divide the equation by $2\cos 6{}^\circ$ therefore we will get,
$\therefore$ L.H.S. (Left Hand Side) $=\sin 6{}^\circ \times \dfrac{2\cos 6{}^\circ }{2\cos 6{}^\circ }\times \cos 12{}^\circ \times \cos 24{}^\circ \times \cos 48{}^\circ$
On rearranging the above equation we will get,
$\therefore$ L.H.S. (Left Hand Side) $=\dfrac{2\times \sin 6{}^\circ \times \cos 6{}^\circ }{2\times \cos 6{}^\circ }\times \cos 12{}^\circ \times \cos 24{}^\circ \times \cos 48{}^\circ$
To proceed further in the solution we should know the formula given below,
Formula:
$\sin 2x=2\times \sin x\times \cos x$ ……………………………… (i)
By using above formula we can write the above equation as,
$\therefore$ L.H.S. (Left Hand Side) $=\dfrac{\sin \left( 2\times 6{}^\circ \right)}{2\cos 6{}^\circ }\times \cos 12{}^\circ \times \cos 24{}^\circ \times \cos 48{}^\circ$
$\therefore$ L.H.S. (Left Hand Side) $=\dfrac{\sin 12{}^\circ }{2\cos 6{}^\circ }\times \cos 12{}^\circ \times \cos 24{}^\circ \times \cos 48{}^\circ$
On rearranging the above equation we will get,
$\therefore$ L.H.S. (Left Hand Side) $=\dfrac{\sin 12{}^\circ \times \cos 12{}^\circ \times \cos 24{}^\circ \times \cos 48{}^\circ }{2\cos 6{}^\circ }$
Multiplying and dividing the above equation by 2 we will get,
$\therefore$ L.H.S. (Left Hand Side) $=\dfrac{\left( 2\times \sin 12{}^\circ \times \cos 12{}^\circ \right)\times \cos 24{}^\circ \times \cos 48{}^\circ }{2\times 2\times \cos 6{}^\circ }$
By using the formula (i) we will get,
$\therefore$ L.H.S. (Left Hand Side) $=\dfrac{\sin 2\left( 12{}^\circ \right)\times \cos 24{}^\circ \times \cos 48{}^\circ }{4\times \cos 6{}^\circ }$
$\therefore$ L.H.S. (Left Hand Side) $=\dfrac{\sin 24{}^\circ \times \cos 24{}^\circ \times \cos 48{}^\circ }{4\times \cos 6{}^\circ }$
Multiplying and dividing the above equation by 2 we will get,
$\therefore$ L.H.S. (Left Hand Side) $=\dfrac{\left( 2\times \sin 24{}^\circ \times \cos 24{}^\circ \right)\times \cos 48{}^\circ }{2\times 4\times \cos 6{}^\circ }$
By using the formula (i) we will get,
$\therefore$ L.H.S. (Left Hand Side) $=\dfrac{\sin 2\left( 24{}^\circ \right)\times \cos 48{}^\circ }{8\times \cos 6{}^\circ }$
$\therefore$ L.H.S. (Left Hand Side) $=\dfrac{\sin 48{}^\circ \times \cos 48{}^\circ }{8\times \cos 6{}^\circ }$
Multiplying and dividing the above equation by 2 we will get,
$\therefore$ L.H.S. (Left Hand Side) $=\dfrac{2\times \sin 48{}^\circ \times \cos 48{}^\circ }{2\times 8\times \cos 6{}^\circ }$
By using the formula (i) we will get,
$\therefore$ L.H.S. (Left Hand Side) $=\dfrac{\sin 2\left( 48{}^\circ \right)}{16\times \cos 6{}^\circ }$
$\therefore$ L.H.S. (Left Hand Side) $=\dfrac{\sin 96{}^\circ }{16\times \cos 6{}^\circ }$
Above equation can also be written as,
$\therefore$ L.H.S. (Left Hand Side) $=\dfrac{\sin \left( 90{}^\circ +6{}^\circ \right)}{16\times \cos 6{}^\circ }$
To proceed further we should know the formula given below,
Formula:
$\sin \left( 90+\theta \right)=\cos \theta$
By using above formula we can write the above equation as,
$\therefore$ L.H.S. (Left Hand Side) $=\dfrac{\cos 6{}^\circ }{16\times \cos 6{}^\circ }$
$\therefore$ L.H.S. (Left Hand Side) $=\dfrac{1}{16}$
$\therefore$ L.H.S. (Left Hand Side) = R.H.S. (Right Hand Side)
$\therefore$ $\sin 6{}^\circ \sin 42{}^\circ \sin 66{}^\circ \sin 78{}^\circ =\dfrac{1}{16}$
Hence proved.

Note: Do not use the formula of ‘2 sinC sinD’ as by using this formula the solution will become very much lengthy and difficult to solve.