Question

Find the value of the trigonometric expression $\sin {50^0} - \sin {70^0} + \sin {10^0}$=
$
  (a){\text{ 0}} \\
  (b){\text{ 1}} \\
  (c){\text{ 2}} \\
  (d){\text{ 3}} \\
$

Answer 1

Hint: In this question we have to find the value of the given trigonometric expression. Use the basic trigonometric identities like $\sin C - \sin D = 2\sin \left( {\dfrac{{C - D}}{2}} \right)\cos \left( {\dfrac{{C + D}}{2}} \right)$and others. This will help in getting the right answer.

Complete step-by-step answer:

Given trigonometric equation is

$\sin {50^0} - \sin {70^0} + \sin {10^0}$

Now as we know the identity $\sin C - \sin D = 2\sin \left( {\dfrac{{C - D}}{2}} \right)\cos \left( {\dfrac{{C + D}}{2}} \right)$ so use this property in above equation we have,

$ \Rightarrow \sin {50^0} - \sin {70^0} + \sin {10^0} = 2\sin \left( {\dfrac{{{{50}^0} -

{{70}^0}}}{2}} \right)\cos \left( {\dfrac{{{{50}^0} + {{70}^0}}}{2}} \right) + \sin {10^0}$

$ \Rightarrow 2\sin \left( { - {{10}^0}} \right)\cos \left( {{{60}^0}} \right) + \sin {10^0}$

Now as we know that the value of $\cos {60^0} = \dfrac{1}{2}$ and the property $\sin \left( {

- \theta } \right) = - \sin \theta $ so use these in above equation and simplify we have,

$ \Rightarrow \sin {50^0} - \sin {70^0} + \sin {10^0} = - 2\sin {10^0}\left( {\dfrac{1}{2}}

\right) + \sin {10^0} = - \sin {10^0} + \sin {10^0} = 0$

So, the answer of the equation is 0.

Hence option (A) is correct.

Note: Whenever we face such types of problems the key concept is simply to have the basic understanding of the trigonometric identities some of them have been mentioned above. This will help in getting on the right track to get the answer.