Question

Find the value of $ \cos 1{}^\circ \cos 2{}^\circ \cos 3{}^\circ ...\cos 180{}^\circ $ .
A. 1
B. -1
C. 0
D. None of these.

Answer 1

Hint: We know that $ \cos 90{}^\circ =0 $. Do not try to simplify by using any formula of "Product to Sum" type. If we multiply anything with 0 the result will also come 0.

Complete step-by-step answer:
We know that the value of $ \cos \theta $ lies between -1 and 1.
Recall that $ \cos 0{}^\circ =1 $ , $ \cos 90{}^\circ =0 $ and $ \cos 180{}^\circ =-1 $ .
Since $ \cos 90{}^\circ =0 $ is one of the terms in the product $ \cos 1{}^\circ \cos 2{}^\circ \cos 3{}^\circ ...\cos 180{}^\circ $ , the final product will be 0.
Hence, the correct answer is C. 0.

Note:The value of $ \sin \theta $ and $ \cos \theta $ lies between -1 and 1.
Remember the trigonometric formula to solve these questions easily.
Sum-Product formula:
 $ \sin 2A+\sin 2B=2\sin (A+B)\cos (A-B) $
 $ \sin 2A-\sin 2B=2\cos (A+B)\sin (A-B) $
 $ \cos 2A+\cos 2B=2\cos (A+B)\cos (A-B) $
 $ \cos 2A-\cos 2B=-2\sin (A+B)\sin (A-B) $
Trigonometric Ratios for Allied Angles:
 $ \sin \left( -\theta \right)=-\sin \theta $ $ \cos \left( -\theta \right)=\cos \theta $
 $ \sin \left( 2n\pi +\theta \right)=\sin \theta $ $ \cos \left( 2n\pi +\theta \right)=\cos \theta $
 $ \sin \left( n\pi +\theta \right)={{\left( -1 \right)}^{n}}\sin \theta $ $ \cos \left( n\pi +\theta \right)={{\left( -1 \right)}^{n}}\cos \theta $
 $ \sin \left[ (2n+1)\dfrac{\pi }{2}+\theta \right]={{(-1)}^{n}}\cos \theta $ $ \cos \left[ (2n+1)\dfrac{\pi }{2}+\theta \right]={{(-1)}^{n}}\left( -\sin \theta \right) $