Question

Find $ \sin {16^ \circ } $ using $ \sin (A - B) = \sin A\cos B - \cos A\sin B $ ?

Answer 1

Hint: As we know that the above question is related to trigonometry as sine, cosine are the trigonometric ratios. The given trigonometric identity is a basic difference formula of the trigonometric ratio. In the above question we will break down the degree in two different numbers and then apply the given trigonometric identity.

Complete step by step solution:
As per the given question we have $ \sin {16^ \circ } $ .
It can be written as $ \sin 16 = \sin (90 - 74) $ .Here we have $ A = 90 $ and $ B = 74 $ .
So we can write it as $ \sin (90 - 74) = \sin 90\cos 74 - \cos 90\sin 74 $ .
We know the value of $ \sin 90 = 1 $ and $ \cos 90 = 0 $ . The value of $ \cos 74 $ is the same in degrees and radians. To obtain the value of $ {74^ \circ } $ in radian, we multiply $ 74 $ by $ \dfrac{\pi }{{180}} = \dfrac{{37}}{{90}} \times \pi $ .
Therefore the value is $ 0.275639(approx) $ .
By putting the values in the formula we have $ \sin {16^ \circ } = 1 \times 0.27563 - 0 \times \sin 74 $ .
Hence the required value of $ \sin {16^ \circ } $ is $ 0.27563 $ .
So, the correct answer is “ $ 0.27563 $ ”.

Note: Before solving this kind of question we should have the clear concept of trigonometric ratios, identities and their formulas. We should note that in the above solution $ \sin (90 - 74) $ can also be written as $ \cos 74 $ , as the identity says that $ \sin (90 - \theta ) = \cos \theta $ . It gives us the same value as in the above solution. WE should know all the identities as there are several ways to solve a question.