Question

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

Answer 1

To find the value of expressions given in trigonometric functions neither in degrees or radians use trigonometric values from table and find the value of the given expression $\cos {1^0}\cos {2^0}\cos {3^0}......\cos {180^0}$.

Complete step by step solution:
The objective of the problem is to find the value of $\cos {1^0}\cos {2^0}\cos {3^0}......\cos {180^0}$
Actually it is so hard to remember the value of $\cos {1^0},\cos {2^0},\cos {3^0},$ and so on
The values of $\cos {1^0},\cos {2^0},\cos {3^0},$….. are obtained by using logarithms table book or by using a calculator.
But in this problem no need to follow this procedure we think of an alternative method which has become easier and simple. For this we use the algebraic property of 0.
The product of numbers with zero is zero, that is $a \times 0 = 0$
In the expression $\cos {1^0}\cos {2^0}\cos {3^0}......\cos {189^0}$ there will be $\cos {90^0}$
We know that the value of $\cos {90^0}$ is zero that is $\cos {90^0} = 0$
Hence we write ,
$\cos {1^0}\cos {2^0}\cos {3^0}......\cos {180^0} = \cos {1^0}\cos {2^0}\cos {3^0}...\cos {90^0}...\cos {180^0}$
$= \cos {1^0}\cos {2^0}\cos {3^0}... \times 0 \times ...\cos {180^0}$
$= 0$

Since the product with zero is zero . the value of the given equation is also zero.
Therefore the value of $\cos {1^0}\cos {2^0}\cos {3^0}......\cos {180^0}$ is 0.
The cosine trigonometric function is positive in the first and fourth quadrants in the coordinate axis. And in the remaining coordinates the cosine function is negative .
The values of cosine function are $\cos \,0 = 1\,,\cos \,30 = \dfrac{{\sqrt 3 }}{2},\cos 45 = \dfrac{1}{{\sqrt 2 }},\cos 60 = \dfrac{1}{2},\cos 90 = 0$
Cos decreases from 1 to 0 in the interval $\left[ {0,\dfrac{\pi }{2}} \right]$ and again to -1 to increase to 1. The value of cos is zero at multiples of $\dfrac{\pi }{2}$ and the value is one at even multiples of pi. Similarly the value is -1 at odd multiples of pi.
Hence , the value of the given problem is zero .

So, the correct answer is “Option C”.

Note: The sine trigonometric function is positive in first and second quadrants in the coordinate axis and in the remaining it is negative only. And for the tan trigonometric function it is positive in the first and third quadrants of the coordinate system.