Question

How do you find the exact value of $\cos \left( {3\pi } \right)$ ?

Answer 1

Hint:For solving this very question we will first write the expression in such a way that it will follow the formula given by $\cos \left( {A + B} \right) = \cos A.\cos B - \sin A.\sin B$ . And then substituting the values we will get to the result.

Formula used:
The formula in terms of cosine,
$\cos \left( {A + B} \right) = \cos A.\cos B - \sin A.\sin B$

Complete step by step answer:
So we have the expression given as $\cos \left( {3\pi } \right)$.
And for solving it we will first expand the expression and it can be written as
$ \Rightarrow \cos \left( {2\pi + \pi } \right)$
And as we know the formula $\cos \left( {A + B} \right) = \cos A.\cos B - \sin A.\sin B$ so by comparing the LHS of the equation with the above expression we have the values of constants will be as $A = 2\pi ,B = \pi $.
Therefore, on substituting the values, we will get the equation as
$ \Rightarrow \cos \left( {2\pi + \pi } \right) = \cos 2\pi \cdot \cos \pi - \sin 2\pi \cdot \sin \pi $
And as we know the value of $\sin \pi = 0$ and they're even multiple will also be the same. Whereas the value of $\cos \pi = 1$ and for the even multiple the sign will keep changing at the interval.
Therefore, on substituting the values, we will get the equation as
$ \Rightarrow \cos \left( {2\pi + \pi } \right) = 1 \cdot \left( { - 1} \right) - 0 \cdot 0$
And on solving it we will get
$ \Rightarrow \cos \left( {2\pi + \pi } \right) = - 1$

Therefore, the exact value of $\cos \left( {3\pi } \right)$ will be equal to $ - 1$.

Note: For solving such types of questions we need to memorize and remember the formula used in it. And also we have to remember that the odd multiple of cosine function will be negative, the even multiple will always be positive.