How do you find the exact functional value $ \sin (60^\circ + 45^\circ ) $ using the cosine sum or difference identity?

Hint: All the trigonometric functions are related to each other through several identities. Bigger angles are expressed as a sum of smaller angles and their values are found using the identities. In the given question, we have to find the sine of $ 60^\circ + 45^\circ $ by using the sum or difference identity, so first; we will apply the appropriate identity and then plug in the known values to get the correct answer.

Complete step-by-step answer:
Given,
 $ \sin (60^\circ + 45^\circ ) $
We know that –
 $
  \sin (a + b) = \sin a\cos b + \cos a\sin b \\
   \Rightarrow \sin (60^\circ + 45^\circ ) = \sin 60^\circ \cos 45^\circ + \cos 60^\circ \sin 45^\circ \\
   \Rightarrow \sin (60^\circ + 45^\circ ) = \dfrac{{\sqrt 3 }}{2} \times \dfrac{1}{{\sqrt 2 }} + \dfrac{1}{2} \times \dfrac{1}{{\sqrt 2 }} \\
   \Rightarrow \sin (60^\circ + 45^\circ ) = \dfrac{{\sqrt 3 + 1}}{{2\sqrt 2 }} \\
  $
Hence, the exact functional value of $ \sin (60^\circ + 45^\circ ) $ is $ \dfrac{{\sqrt 3 + 1}}{{2\sqrt 2 }} $ .
So, the correct answer is “ $ \dfrac{{\sqrt 3 + 1}}{{2\sqrt 2 }} $ ”.

Note: We know the value of the sine and cosine function when the angle lies between 0 and $ \dfrac{\pi }{2} $ .So we simply plugged the values of $ 60^\circ $ and $ 45^\circ $ . We have applied the identity that the sine of the sum of two angles a and b is equal to the sum of the product of the sine of angle a and cosine of angle b and the product of the cosine of angle a and the sine of angle b that is $ \sin (a + b) = \sin a\cos b - \cos a\sin b $. Similarly, $ \sin (a - b) = \sin a\cos b - \cos a\sin b $ . Many such identities can be used to solve similar questions.