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Question: Find the value of \( \sin {10^ \circ } + \sin {130^ \circ } - \sin {110^ \circ } \) ?...

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

Explanation

Solution

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=2sin⁡(A+B2)cos⁡(A−B2)\sin A + \sin B = 2\sin \left( {\dfrac{{A + B}}{2}} \right)\cos \left( {\dfrac{{A - B}}{2}} \right)
sin⁡A−sin⁡B=2cos⁡(A+B2)sin⁡(A−B2)\sin A - \sin B = 2\cos \left( {\dfrac{{A + B}}{2}} \right)\sin \left( {\dfrac{{A - B}}{2}} \right)
cos⁡(−x)=cos⁡x\cos \left( { - x} \right) = \cos x
cos⁡(90∘)=0\cos \left( {{{90}^ \circ }} \right) = 0
cos⁡(60∘)=12\cos \left( {{{60}^ \circ }} \right) = \dfrac{1}{2}

Complete step by step solution:
We are given the equation sin⁡10∘+sin⁡130∘−sin⁡110∘\sin {10^ \circ } + \sin {130^ \circ } - \sin {110^ \circ } .
Taking two operands at a time suppose sin⁡10∘\sin {10^ \circ } and sin⁡130∘\sin {130^ \circ } , pairing them in a parenthesis, and we get:
(sin⁡10∘+sin⁡130∘)−sin⁡110∘\left( {\sin {{10}^ \circ } + \sin {{130}^ \circ }} \right) - \sin {110^ \circ } …………….(1)
From the trigonometric identities, we know that
sin⁡A+sin⁡B=2sin⁡(A+B2)cos⁡(A−B2)\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∘+sin⁡130∘\sin {10^ \circ } + \sin {130^ \circ } with sin⁡A+sin⁡B\sin A + \sin B , we get the value of:
A=10∘A = {10^ \circ }
B=130∘B = {130^ \circ }
Substituting these values in the formula sin⁡A+sin⁡B=2sin⁡(A+B2)cos⁡(A−B2)\sin A + \sin B = 2\sin \left( {\dfrac{{A + B}}{2}} \right)\cos \left( {\dfrac{{A - B}}{2}} \right) :
sin⁡10∘+sin⁡130∘=2sin⁡(10∘+130∘2)cos⁡(10∘−130∘2)\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∘+sin⁡130∘=2sin⁡(140∘2)cos⁡(−120∘2)\sin {10^ \circ } + \sin {130^ \circ } = 2\sin \left( {\dfrac{{{{140}^ \circ }}}{2}} \right)\cos \left( {\dfrac{{ - {{120}^ \circ }}}{2}} \right)
sin⁡10∘+sin⁡130∘=2sin⁡(70∘)cos⁡(−60∘)\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⁡(−x)=cos⁡x\cos \left( { - x} \right) = \cos x .So from this we get cos⁡(−60∘)=cos⁡(60∘)\cos \left( { - {{60}^ \circ }} \right) = \cos \left( {{{60}^ \circ }} \right) .
Substituting this value in equation (2), we get:
sin⁡10∘+sin⁡130∘=2sin⁡(70∘)cos⁡(60∘)\sin {10^ \circ } + \sin {130^ \circ } = 2\sin \left( {{{70}^ \circ }} \right)\cos \left( {{{60}^ \circ }} \right)
Since, we know that cos⁡(60∘)=12\cos \left( {{{60}^ \circ }} \right) = \dfrac{1}{2} , so substituting this in the above value, we get:
sin⁡10∘+sin⁡130∘=2sin⁡(70∘)×12\sin {10^ \circ } + \sin {130^ \circ } = 2\sin \left( {{{70}^ \circ }} \right) \times \dfrac{1}{2}
Cancelling out the 2′s2's we get:
sin⁡10∘+sin⁡130∘=sin⁡(70∘)\sin {10^ \circ } + \sin {130^ \circ } = \sin \left( {{{70}^ \circ }} \right)
Now, substituting this value in the main equation that is equation (1):
(sin⁡10∘+sin⁡130∘)−sin⁡110∘\left( {\sin {{10}^ \circ } + \sin {{130}^ \circ }} \right) - \sin {110^ \circ }
⇒sin⁡70∘−sin⁡110∘\Rightarrow \sin {70^ \circ } - \sin {110^ \circ } …………..(3)
From the trigonometric formulas, we know that
sin⁡A−sin⁡B=2cos⁡(A+B2)sin⁡(A−B2)\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∘−sin⁡110∘\sin {70^ \circ } - \sin {110^ \circ } with sin⁡A−sin⁡B\sin A - \sin B , we get the value of:
A=70∘A = {70^ \circ }
B=110∘B = {110^ \circ }
Substituting these values in the formula
sin⁡A−sin⁡B=2cos⁡(A+B2)sin⁡(A−B2)\sin A - \sin B = 2\cos \left( {\dfrac{{A + B}}{2}} \right)\sin \left( {\dfrac{{A - B}}{2}} \right) :
sin⁡70∘+sin⁡110∘=2cos⁡(70∘+110∘2)sin⁡(70∘−110∘2)\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∘+sin⁡110∘=2cos⁡(180∘2)sin⁡(−40∘2)\sin {70^ \circ } + \sin {110^ \circ } = 2\cos \left( {\dfrac{{{{180}^ \circ }}}{2}} \right)\sin \left( {\dfrac{{ - {{40}^ \circ }}}{2}} \right)
sin⁡70∘+sin⁡110∘=2cos⁡(90∘)sin⁡(−20∘)\sin {70^ \circ } + \sin {110^ \circ } = 2\cos \left( {{{90}^ \circ }} \right)\sin \left( { - {{20}^ \circ }} \right)
Since, we know that cos⁡(90∘)=0\cos \left( {{{90}^ \circ }} \right) = 0 , so substituting this in the above value, we get:
sin⁡70∘+sin⁡110∘=2×0×sin⁡(−20∘)\sin {70^ \circ } + \sin {110^ \circ } = 2 \times 0 \times \sin \left( { - {{20}^ \circ }} \right)
Solving the equation:
sin⁡70∘+sin⁡110∘=0\sin {70^ \circ } + \sin {110^ \circ } = 0
Therefore, the value of sin⁡10∘+sin⁡130∘−sin⁡110∘\sin {10^ \circ } + \sin {130^ \circ } - \sin {110^ \circ } is 00 .
So, the correct answer is “0”.

Note : It’s not compulsory to take the first two operands only, we could have taken sin⁡10∘,sin⁡110∘\sin {10^ \circ },\sin {110^ \circ } in the first step, then further with the third as (sin⁡10∘−sin⁡110∘)+sin⁡130∘\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