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Question: If \[\sin {24^ \circ } + \sin {32^ \circ } + \sin {204^ \circ } + \sin {212^ \circ }\] is equals to ...

If sin⁡24∘+sin⁡32∘+sin⁡204∘+sin⁡212∘\sin {24^ \circ } + \sin {32^ \circ } + \sin {204^ \circ } + \sin {212^ \circ } is equals to
A.-1
B.1
C.0
D.2

Explanation

Solution

Hint : Given are the angles of trigonometric functions. If we observed then the last two angles are angles that can be formed as 180∘+24∘{180^ \circ } + {24^ \circ } and 180∘+32∘{180^ \circ } + {32^ \circ }. Now we can change that angle and rewrite it and then use the property of angles to solve further.

Complete step-by-step answer :
Given the question is,
sin⁡24∘+sin⁡32∘+sin⁡204∘+sin⁡212∘\sin {24^ \circ } + \sin {32^ \circ } + \sin {204^ \circ } + \sin {212^ \circ }
We can rewrite the angles as,
=sin⁡24∘+sin⁡32∘+sin⁡(180∘+24∘)+sin⁡(180∘+32∘)= \sin {24^ \circ } + \sin {32^ \circ } + \sin \left( {{{180}^ \circ } + {{24}^ \circ }} \right) + \sin \left( {{{180}^ \circ } + {{32}^ \circ }} \right)
Now we know that, sin⁡(180∘+θ)=−sin⁡θ\sin \left( {{{180}^ \circ } + \theta } \right) = - \sin \theta
So we can write the above equation as,
=sin⁡24∘+sin⁡32∘−sin⁡(24∘)−sin⁡(32∘)= \sin {24^ \circ } + \sin {32^ \circ } - \sin \left( {{{24}^ \circ }} \right) - \sin \left( {{{32}^ \circ }} \right)
Now since the first two angles and last two angles are the same we will simply cancel them or the answer will be zero.
=0= 0
So , sin⁡24∘+sin⁡32∘+sin⁡204∘+sin⁡212∘=0\sin {24^ \circ } + \sin {32^ \circ } + \sin {204^ \circ } + \sin {212^ \circ } = 0
So the correct option is C.
So, the correct answer is “Option C”.

Note : Note that we need not to find the value of the angle in this case or no need of using any trigonometric sum and difference formula. So simply first observe the angle and then write it in the form of either 180∘±θ{180^ \circ } \pm \theta or 90∘±θ{90^ \circ } \pm \theta . This is the way to solve it!