Prove that: cos 7° cos 14° cos 28° cos 56° = sin 68° / 16 cos 83°
Question
Prove that \[ \cos 7^\circ \cos 14^\circ \cos 28^\circ \cos 56^\circ = \frac{\sin 68^\circ}{16\cos 83^\circ} \]
Solution
We use the identity \[ 2\sin \theta \cos \theta = \sin 2\theta \]
Start with \[ \cos 7^\circ \cos 14^\circ \cos 28^\circ \cos 56^\circ \]
Multiply and divide by \[ \sin 7^\circ \]
\[ = \frac{ \sin 7^\circ \cos 7^\circ \cos 14^\circ \cos 28^\circ \cos 56^\circ } {\sin 7^\circ} \]
Using \[ 2\sin A\cos A=\sin 2A \]
\[ = \frac{ \frac{1}{2}\sin 14^\circ \cos 14^\circ \cos 28^\circ \cos 56^\circ } {\sin 7^\circ} \]
Again, \[ 2\sin 14^\circ \cos 14^\circ = \sin 28^\circ \]
\[ = \frac{ \frac{1}{4}\sin 28^\circ \cos 28^\circ \cos 56^\circ } {\sin 7^\circ} \]
Again, \[ 2\sin 28^\circ \cos 28^\circ = \sin 56^\circ \]
\[ = \frac{ \frac{1}{8}\sin 56^\circ \cos 56^\circ } {\sin 7^\circ} \]
Again, \[ 2\sin 56^\circ \cos 56^\circ = \sin 112^\circ \]
\[ = \frac{ \frac{1}{16}\sin 112^\circ } {\sin 7^\circ} \]
Since \[ \sin 112^\circ = \sin(180^\circ-112^\circ) \]
\[ = \sin 68^\circ \]
and \[ \sin 7^\circ = \cos 83^\circ \]
Therefore,
\[ \cos 7^\circ \cos 14^\circ \cos 28^\circ \cos 56^\circ = \frac{\sin 68^\circ}{16\cos 83^\circ} \]
Final Answer
\[ \boxed{ \cos 7^\circ \cos 14^\circ \cos 28^\circ \cos 56^\circ = \frac{\sin 68^\circ}{16\cos 83^\circ} } \]