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Question: \[\frac{1}{4}\left\lbrack \sqrt{3}\cos 23{^\circ} - \sin 23{^\circ} \right\rbrack =\]...

14[3cos⁡23∘−sin⁡23∘]=\frac{1}{4}\left\lbrack \sqrt{3}\cos 23{^\circ} - \sin 23{^\circ} \right\rbrack =

A

cos⁡43∘\cos 43{^\circ}

B

cos⁡7∘\cos 7{^\circ}

C

cos⁡53∘\cos 53{^\circ}

D

None of these

Answer

None of these

Explanation

Solution

14{3cos⁡23o−sin⁡23o}\frac{1}{4}\{\sqrt{3}\cos 23^{o} - \sin 23^{o}\}

=12{cos⁡30ocos⁡23o−sin⁡30osin⁡23o}= \frac{1}{2}\{\cos 30^{o}\cos 23^{o} - \sin 30^{o}\sin 23^{o}\}

=12cos⁡(30o+23o)=12cos⁡53o.= \frac{1}{2}\cos(30^{o} + 23^{o}) = \frac{1}{2}\cos 53^{o}.