
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
544.2k+ views
Hint: 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.
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.
Recently Updated Pages
Master Class 11 Computer Science: Engaging Questions & Answers for Success

Master Class 11 Business Studies: Engaging Questions & Answers for Success

Master Class 11 Economics: Engaging Questions & Answers for Success

Master Class 11 English: Engaging Questions & Answers for Success

Master Class 11 Maths: Engaging Questions & Answers for Success

Master Class 11 Biology: Engaging Questions & Answers for Success

Trending doubts
One Metric ton is equal to kg A 10000 B 1000 C 100 class 11 physics CBSE

There are 720 permutations of the digits 1 2 3 4 5 class 11 maths CBSE

Discuss the various forms of bacteria class 11 biology CBSE

Draw a diagram of a plant cell and label at least eight class 11 biology CBSE

State the laws of reflection of light

Explain zero factorial class 11 maths CBSE

