Question

Find the value of $ \sin {10^ \circ } + \sin {130^ \circ } - \sin {110^ \circ } $ ?

Answer 1

Hint: In order to solve the above equation, there is only use of trigonometric identities. Trigonometric identities are some fixed formulas obtained to solve the trigonometric equations. Take two operands at a time, apply the formula described below and then solve it to get another operand, then continue the process with the left operand.
Formula used:
 $ \sin A + \sin B = 2\sin \left( {\dfrac{{A + B}}{2}} \right)\cos \left( {\dfrac{{A - B}}{2}} \right) $
 $ \sin A - \sin B = 2\cos \left( {\dfrac{{A + B}}{2}} \right)\sin \left( {\dfrac{{A - B}}{2}} \right) $
 $ \cos \left( { - x} \right) = \cos x $
 $ \cos \left( {{{90}^ \circ }} \right) = 0 $
 $ \cos \left( {{{60}^ \circ }} \right) = \dfrac{1}{2} $

Complete step by step solution:
We are given the equation $ \sin {10^ \circ } + \sin {130^ \circ } - \sin {110^ \circ } $ .
Taking two operands at a time suppose $ \sin {10^ \circ } $ and $ \sin {130^ \circ } $ , pairing them in a parenthesis, and we get:
\[\left( {\sin {{10}^ \circ } + \sin {{130}^ \circ }} \right) - \sin {110^ \circ }\] …………….(1)
From the trigonometric identities, we know that
  $ \sin A + \sin B = 2\sin \left( {\dfrac{{A + B}}{2}} \right)\cos \left( {\dfrac{{A - B}}{2}} \right) $ .
Comparing the value inside the brackets
  $ \sin {10^ \circ } + \sin {130^ \circ } $ with $ \sin A + \sin B $ , we get the value of:
 $ A = {10^ \circ } $
 $ B = {130^ \circ } $
Substituting these values in the formula $ \sin A + \sin B = 2\sin \left( {\dfrac{{A + B}}{2}} \right)\cos \left( {\dfrac{{A - B}}{2}} \right) $ :
 $ \sin {10^ \circ } + \sin {130^ \circ } = 2\sin \left( {\dfrac{{{{10}^ \circ } + {{130}^ \circ }}}{2}} \right)\cos \left( {\dfrac{{{{10}^ \circ } - {{130}^ \circ }}}{2}} \right) $
Solving the values inside the parenthesis, we get:
 $ \sin {10^ \circ } + \sin {130^ \circ } = 2\sin \left( {\dfrac{{{{140}^ \circ }}}{2}} \right)\cos \left( {\dfrac{{ - {{120}^ \circ }}}{2}} \right) $
 $ \sin {10^ \circ } + \sin {130^ \circ } = 2\sin \left( {{{70}^ \circ }} \right)\cos \left( { - {{60}^ \circ }} \right) $ ……….(2)
Since, we know that cosine is an even function that gives $ \cos \left( { - x} \right) = \cos x $ .So from this we get $ \cos \left( { - {{60}^ \circ }} \right) = \cos \left( {{{60}^ \circ }} \right) $ .
Substituting this value in equation (2), we get:
 $ \sin {10^ \circ } + \sin {130^ \circ } = 2\sin \left( {{{70}^ \circ }} \right)\cos \left( {{{60}^ \circ }} \right) $
Since, we know that $ \cos \left( {{{60}^ \circ }} \right) = \dfrac{1}{2} $ , so substituting this in the above value, we get:
 $ \sin {10^ \circ } + \sin {130^ \circ } = 2\sin \left( {{{70}^ \circ }} \right) \times \dfrac{1}{2} $
Cancelling out the $ 2's $ we get:
 $ \sin {10^ \circ } + \sin {130^ \circ } = \sin \left( {{{70}^ \circ }} \right) $
Now, substituting this value in the main equation that is equation (1):
\[\left( {\sin {{10}^ \circ } + \sin {{130}^ \circ }} \right) - \sin {110^ \circ }\]
\[ \Rightarrow \sin {70^ \circ } - \sin {110^ \circ }\] …………..(3)
From the trigonometric formulas, we know that
  $ \sin A - \sin B = 2\cos \left( {\dfrac{{A + B}}{2}} \right)\sin \left( {\dfrac{{A - B}}{2}} \right) $ .
Comparing the equation (3) , $ \sin {70^ \circ } - \sin {110^ \circ } $ with $ \sin A - \sin B $ , we get the value of:
 $ A = {70^ \circ } $
 $ B = {110^ \circ } $
Substituting these values in the formula
 $ \sin A - \sin B = 2\cos \left( {\dfrac{{A + B}}{2}} \right)\sin \left( {\dfrac{{A - B}}{2}} \right) $ :
 $ \sin {70^ \circ } + \sin {110^ \circ } = 2\cos \left( {\dfrac{{{{70}^ \circ } + {{110}^ \circ }}}{2}} \right)\sin \left( {\dfrac{{{{70}^ \circ } - {{110}^ \circ }}}{2}} \right) $
Solving the values inside the parenthesis, we get:
 $ \sin {70^ \circ } + \sin {110^ \circ } = 2\cos \left( {\dfrac{{{{180}^ \circ }}}{2}} \right)\sin \left( {\dfrac{{ - {{40}^ \circ }}}{2}} \right) $
 $ \sin {70^ \circ } + \sin {110^ \circ } = 2\cos \left( {{{90}^ \circ }} \right)\sin \left( { - {{20}^ \circ }} \right) $
Since, we know that $ \cos \left( {{{90}^ \circ }} \right) = 0 $ , so substituting this in the above value, we get:
 $ \sin {70^ \circ } + \sin {110^ \circ } = 2 \times 0 \times \sin \left( { - {{20}^ \circ }} \right) $
Solving the equation:
 $ \sin {70^ \circ } + \sin {110^ \circ } = 0 $
Therefore, the value of $ \sin {10^ \circ } + \sin {130^ \circ } - \sin {110^ \circ } $ is $ 0 $ .
So, the correct answer is “0”.

Note: It’s not compulsory to take the first two operands only, we could have taken $ \sin {10^ \circ },\sin {110^ \circ } $ in the first step, then further with the third as $ \left( {\sin {{10}^ \circ } - \sin {{110}^ \circ }} \right) + \sin {130^ \circ } $ , it would have resulted in the same answer.
It’s important to remember the correct trigonometric identities and should be placed in their appropriate value in order to avoid mistakes