Question

How do I find the exact value of \[\cos 240\]?

Answer 1

Hint: In the given question, we have been asked to find the value of a trigonometric function. Now, the argument of the given trigonometric function is not in the range of the known values of the trigonometric functions as given in the standard table, in which values lie from \[0\] to \[\pi /2\]. But we can calculate that by breaking the given trigonometric function by writing the angle of the function as a sum of two angles, such that when we apply the sum formula of the given trigonometric function, we get the individual trigonometric functions whose value is known.

Formula Used:
We are going to use the sum formula of cosecant, which is:
\[\cos \left( {a + b} \right) = \cos \left( a \right)\cos \left( b \right) - \sin \left( a \right)\sin \left( b \right)\].

Complete step by step answer:
We have to find the value of \[\cos 240^\circ \].
We can represent the angle used in the trigonometric function as \[240^\circ = 180^\circ + 60^\circ \], so, we have:
\[\cos 240^\circ = \cos \left( {180 + 60} \right)\]
Applying the formula, we get:
\[\cos 240^\circ = \cos 180\cos 60 - \sin 60\sin 180\]
We know, \[\cos 180^\circ = - 1,{\rm{ }}\cos 60^\circ = \dfrac{1}{2}\] and the second part does not matter as \[\sin 180^\circ = 0\], hence, we get:

\[\cos 240^\circ = - 1 \times \dfrac{1}{2} = - \dfrac{1}{2}\]

Additional Information:
In the given question, we applied the sum formula. But, if the sum formula does not have any value which is in the known range, we can apply the difference formula for the given trigonometric function and then solve for the angle given in the question.

Note: In this question, we applied the sum formula because the given argument can be expressed as the sum of two arguments whose individual value is known. But it is always better if we know the exact identity of the relation between two trigonometric functions when they are expressed as the sum or difference of two angles, out of which one angle is always a multiple of \[90^\circ \].