Question

How do you find the exact functional value of $ \cos (285^\circ ) $ using the cosine sum or difference identity?

Answer 1

Hint: The relation between the sides of a right-angled triangle that is the base, the perpendicular and the hypotenuse is studied using trigonometry. All the trigonometric functions are interrelated with each other through some identities and have great importance in mathematics and real-life too. To find the trigonometric value of large angles, there are many ways. In the given question, we have to find the cosine of 285 degrees by using the sum or difference identity, so first, we will write 285 degrees as the sum of two angles such that their value is already known or can be found easily.

Complete step-by-step answer:
As the trigonometric values of 135 and 150 are easy to find, 285 degrees can be written as a sum of 135 and 150 –
 $ \cos (285^\circ ) = \cos (135^\circ + 150^\circ ) $
We know that –
 $
  \cos (A + B) = \cos A\cos B - \sin A\sin B \\
   \Rightarrow \cos (135 + 150) = \cos 135\cos 150 - \sin 135\sin 180 \\
   \;
  $
Now,
 $
  \cos 135^\circ = \cos (180^\circ - 45^\circ ) = - \cos 45^\circ = - \dfrac{1}{{\sqrt 2 }} \\
  \sin 135^\circ = \sin (180^\circ - 45^\circ ) = \sin 45^\circ = \dfrac{1}{{\sqrt 2 }} \\
  \cos 150^\circ = \cos (180^\circ - 30^\circ ) = - \cos 30^\circ = - \dfrac{{\sqrt 3 }}{2} \\
  \sin 150^\circ = \sin (180^\circ - 30^\circ ) = \sin 30^\circ = \dfrac{1}{2} \;
  $
Using the above values in the obtained equation, we get –
 $
  \cos 285^\circ = - \dfrac{1}{{\sqrt 2 }} \times - \dfrac{{\sqrt 3 }}{2} - \dfrac{1}{{\sqrt 2 }} \times \dfrac{1}{2} \\
  \cos 285^\circ = \dfrac{{\sqrt 3 }}{{2\sqrt 2 }} - \dfrac{1}{{2\sqrt 2 }} \\
   \Rightarrow \cos 285^\circ = \dfrac{{\sqrt 3 - 1}}{{2\sqrt 2 }} \;
  $
Hence, the exact functional value of $ \cos 285^\circ $ is $ \dfrac{{\sqrt 3 - 1}}{{2\sqrt 2 }} $ .
So, the correct answer is “$ \dfrac{{\sqrt 3 - 1}}{{2\sqrt 2 }} $ ”.

Note: All the trigonometric functions can be represented on a graph paper; in the first quadrant, all of the functions have a positive value. The cosine function is negative in the second quadrant while the sine function is positive in the second quadrant that’s why $ \cos (180 - \theta ) = - \cos \theta $ and $ \sin (180 - \theta ) = \sin \theta $ and we also know that the trigonometric functions are periodic. We know the value of the cosine function when the angle lies between 0 and $ \dfrac{\pi }{2} $ . So to find the value of the cosine of the angles greater than $ \dfrac{\pi }{2} $ , we use the periodic property of these functions.