Solveeit Logo
Login

Question

Question: The value of \(\sin {10^ \circ } + \sin {20^ \circ } + \sin {30^ \circ } + .... + \sin {360^ \circ }...

The value of sin⁡10∘+sin⁡20∘+sin⁡30∘+....+sin⁡360∘\sin {10^ \circ } + \sin {20^ \circ } + \sin {30^ \circ } + .... + \sin {360^ \circ } is equal to:
A. 0
B. 1
C. 3\sqrt 3
D. 2

Explanation

Solution

Hint : The trigonometrical ratios of an angle are numerical quantities. Each one of them represents the ratio of the length of one side to another of a right angled triangle. An angle made up of the algebraic sum of two or more angles is called a compound angle.

Complete step-by-step answer :
The series of sine angles extend up to 360° and the other angles can be written with respect to 360° that is in the form of 360°, therefore writing the above question in the form,
⇒sin⁡(360∘−350∘)+sin⁡(360∘−340∘)+sin⁡(360∘−330∘)+......sin⁡(360∘−190∘)+sin⁡(360∘−180∘)..+sin⁡360∘\Rightarrow \sin ({360^ \circ } - {350^ \circ }) + \sin ({360^ \circ } - {340^ \circ }) + \sin ({360^ \circ } - {330^ \circ }) + ......\sin ({360^ \circ } - {190^ \circ }) + \sin ({360^ \circ } - {180^ \circ }).. + \sin {360^ \circ }
where 360° can be written as 2π and we know that sin2π=0 and the compound angle in sine can be written as :
⇒sin⁡(A−B)=sin⁡Acos⁡B−cos⁡Asin⁡B ∴sin⁡2π=0,cos⁡2π=1   \Rightarrow \sin (A - B) = \sin A\cos B - \cos A\sin B \\\ \therefore \sin 2\pi = 0,\cos 2\pi = 1 \;
Therefore sin (360°-350°) can be written as:
⇒sin⁡(2π−350∘)=sin⁡2π×cos⁡350∘−cos⁡2π×sin⁡350∘ ⇒0+sin⁡350∘=sin⁡350∘   \Rightarrow \sin (2\pi - {350^ \circ }) = \sin 2\pi \times \cos {350^ \circ } - \cos 2\pi \times \sin {350^ \circ } \\\ \Rightarrow 0 + \sin {350^ \circ } = \sin {350^ \circ } \;
Similarly all the terms in compound angle of sine can be written as (–sinB). Therefore we can write the above expression as :
⇒−sin⁡(350∘)−sin⁡(340∘)−sin⁡(330∘)−......−sin⁡(190∘)+sin⁡(180∘)..+sin⁡340∘+sin⁡350∘+sin⁡360∘\Rightarrow - \sin ({350^ \circ }) - \sin ({340^ \circ }) - \sin ({330^ \circ }) - ...... - \sin ({190^ \circ }) + \sin ({180^ \circ }).. + \sin {340^ \circ } + \sin {350^ \circ } + \sin {360^ \circ }
Therefore we can observe the terms similar in magnitude but opposite in sign that is why they will cancel each other. Hence all the terms will cancel each other except the middle term and the last term as they have no term opposite to it and are similar in magnitude. The left terms are both positive:
⇒sin⁡180∘+sin⁡360∘ ⇒sin⁡π+sin⁡2π ∴sin⁡π=0,sin⁡2π=0 ⇒0+0=0   \Rightarrow \sin {180^ \circ } + \sin {360^ \circ } \\\ \Rightarrow \sin \pi + \sin 2\pi \\\ \therefore \sin \pi = 0,\sin 2\pi = 0 \\\ \Rightarrow 0 + 0 = 0 \;
Therefore from the above we can conclude that the correct option is A.
So, the correct answer is “Option A”.

Note : We conclude that
Sin(360°+θ)= sinθ
Sin(360°-θ)= -sinθ
Also sine of the general angle of the form nπ+(−1)nθn\pi + {\left( { - 1} \right)^n}\theta will have the same sine as that of angle θ and so on.