Question

The value of\[\;\sin {50^0} - \sin {70^0} + \sin {10^0}\] is:
$
  {\text{(a) 1}} \\
  {\text{(b) 2}} \\
  {\text{(c) - 1}} \\
  {\text{(d) 0}} \\
 $

Answer 1

Hint: Since the angles are not standard ones , we have to use additional identity of trigonometric functions to convert them in multiplication form and therefore also changing the angles , do this and we will get angles in numbers of 60 and 10, now we can easily simplify.

Complete step-by-step answer:

We have the given trigonometric equation as
\[\;\sin {50^0} - \sin {70^0} + \sin {10^0}\] … (1)
Now we know the identity,
\[\sin A - {\text{ }}\sin {\text{ }}B = 2\cos \dfrac{{A + B}}{2}\sin \dfrac{{A - B}}{2}\]
For simplification, we will split the (1) in two parts,
Firstly, considering \[\sin {50^0} - \sin {70^0}\], we can assume
$A = {50^0}$and $B = {70^0}$
Such that the above mentioned identity can be used,
\[\sin {50^0} - {\text{ }}\sin {\text{ 7}}{{\text{0}}^0} = 2\cos \dfrac{{{{50}^0} + {{70}^0}}}{2}\sin \dfrac{{{{50}^0} - {{70}^0}}}{2}\]
\[ = 2\cos \dfrac{{{{120}^0}}}{2}\sin \dfrac{{ - {{20}^0}}}{2}\]
\[ = 2\cos {60^0}\sin ( - {10^0})\]
Now, we know that, $\cos {60^0} = \dfrac{1}{2}$
Therefore, we get
\[ = 2 \times \dfrac{1}{2} \times \sin ( - {10^0})\]
\[ = - \sin {10^0}\]
Now putting this value in (1), we get
\[ \Rightarrow - \sin {10^0} + \sin {10^0} = 0\]
So, the required answer is (d) $0$.

Note: In solving these types of questions, we must have an adequate knowledge of various trigonometric identities to obtain the required solution that is what we have used above in the question and it is also important to choose the correct terms because sometimes choosing random terms for using identities will not lead us anywhere.