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

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:
$\sin \left( -\theta \right)=-\sin \theta$ $\cos \left( -\theta \right)=\cos \theta$
$\sin \left( 2n\pi +\theta \right)=\sin \theta$ $\cos \left( 2n\pi +\theta \right)=\cos \theta$
$\sin \left( n\pi +\theta \right)={{\left( -1 \right)}^{n}}\sin \theta$ $\cos \left( n\pi +\theta \right)={{\left( -1 \right)}^{n}}\cos \theta$
$\sin \left[ (2n+1)\dfrac{\pi }{2}+\theta \right]={{(-1)}^{n}}\cos \theta$ $\cos \left[ (2n+1)\dfrac{\pi }{2}+\theta \right]={{(-1)}^{n}}\left( -\sin \theta \right)$