Question
Question: Find the solution of the equation given below \(\operatorname{Sin} {10^0} + \operatorname{Sin} {2...
Find the solution of the equation given below
Sin100+Sin200+Sin300+............+Sin3600
Solution
Hint: By using the given trigonometric formula which is given below. The terms can be brought in the form of some series after manipulation.
=Sina+Sin(a+d)+Sin(a+2d)+.........+Sin(a+(n−1)d) =Sin(2d)(Sin(a+(n−1)2d)Sin(2nd)
Given that:
=Sin100+Sin200+Sin300+............+Sin3600 =Sin100+Sin(100+100)+Sin(100+2×100)+...........+Sin(100+35×100) …………………………..(1)
As we know thatSina+Sin(a+d)+Sin(a+2d)+.........+Sin(a+(n−1)d)=Sin(2d)(Sin(a+(n−1)2d)Sin(2nd)
By comparing equation (1) with the above formula, we get
Using the values of a, d, n and substituting these values in the above formula, we get
⇒Sin(210)Sin(10+(36−1)210)Sin(236×10) ⇒Sin5Sin(10+35×5)Sin(36×5)
After simplifying further
⇒Sin50Sin(100+1750)Sin(1800)
Since, the value of Sin1800=0
⇒Sin50Sin(1850)×0 ⇒0
Hence, after simplifying the given trigonometric equation the final result is 0.
Note: For these types of problems, remember all important trigonometric identities and the values of trigonometric functions. Also be aware of the concept of quadrants, range and domain of these functions. Solving these types of problems will become simple if you remember all trigonometric expressions.