# Find the solution of the equation given below

$\operatorname{Sin} {10^0} + \operatorname{Sin} {20^0} + \operatorname{Sin} {30^0} + ............ + \operatorname{Sin} {360^0}$

Last updated date: 24th Mar 2023

•

Total views: 308.4k

•

Views today: 2.85k

Answer

Verified

308.4k+ views

Hint: By using the given trigonometric formula which is given below. The terms can be brought in the form of some series after manipulation.

$

= \operatorname{Sin} a + \operatorname{Sin} (a + d) + \operatorname{Sin} (a + 2d) + ......... + \operatorname{Sin} (a + (n - 1)d) \\

= \dfrac{{(\operatorname{Sin} (a + (n - 1)\dfrac{d}{2})\operatorname{Sin} (\dfrac{{nd}}{2})}}{{\operatorname{Sin} \left( {\dfrac{d}{2}} \right)}} \\

$

Given that:

$

= \operatorname{Sin} {10^0} + \operatorname{Sin} {20^0} + \operatorname{Sin} {30^0} + ............ + \operatorname{Sin} {360^0} \\

= \operatorname{Sin} {10^0} + \operatorname{Sin} ({10^0} + {10^0}) + \operatorname{Sin} ({10^0} + 2 \times {10^0}) + ........... + \operatorname{Sin} ({10^0} + 35 \times {10^0}) \\

$ …………………………..(1)

As we know that$\left[ {\operatorname{Sin} a + \operatorname{Sin} (a + d) + \operatorname{Sin} (a + 2d) + ......... + \operatorname{Sin} (a + (n - 1)d) = \dfrac{{(\operatorname{Sin} (a + (n - 1)\dfrac{d}{2})\operatorname{Sin} (\dfrac{{nd}}{2})}}{{\operatorname{Sin} \left( {\dfrac{d}{2}} \right)}}} \right]$

By comparing equation (1) with the above formula, we get

\[

a = {10^0},d = {10^0} \\

n - 1 = 35 \\

n = 36 \\

\]

Using the values of a, d, n and substituting these values in the above formula, we get

$

\Rightarrow \dfrac{{\operatorname{Sin} (10 + (36 - 1)\dfrac{{10}}{2})\operatorname{Sin} (\dfrac{{36 \times 10}}{2})}}{{\operatorname{Sin} (\dfrac{{10}}{2})}} \\

\Rightarrow \dfrac{{\operatorname{Sin} (10 + 35 \times 5)\operatorname{Sin} (36 \times 5)}}{{\operatorname{Sin} 5}} \\

$

After simplifying further

$ \Rightarrow \dfrac{{\operatorname{Sin} ({{10}^0} + {{175}^0})\operatorname{Sin} ({{180}^0})}}{{\operatorname{Sin} {5^0}}}$

Since, the value of $\operatorname{Sin} {180^0} = 0$

$

\Rightarrow \dfrac{{\operatorname{Sin} ({{185}^0}) \times 0}}{{\operatorname{Sin} {5^0}}} \\

\Rightarrow 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.

$

= \operatorname{Sin} a + \operatorname{Sin} (a + d) + \operatorname{Sin} (a + 2d) + ......... + \operatorname{Sin} (a + (n - 1)d) \\

= \dfrac{{(\operatorname{Sin} (a + (n - 1)\dfrac{d}{2})\operatorname{Sin} (\dfrac{{nd}}{2})}}{{\operatorname{Sin} \left( {\dfrac{d}{2}} \right)}} \\

$

Given that:

$

= \operatorname{Sin} {10^0} + \operatorname{Sin} {20^0} + \operatorname{Sin} {30^0} + ............ + \operatorname{Sin} {360^0} \\

= \operatorname{Sin} {10^0} + \operatorname{Sin} ({10^0} + {10^0}) + \operatorname{Sin} ({10^0} + 2 \times {10^0}) + ........... + \operatorname{Sin} ({10^0} + 35 \times {10^0}) \\

$ …………………………..(1)

As we know that$\left[ {\operatorname{Sin} a + \operatorname{Sin} (a + d) + \operatorname{Sin} (a + 2d) + ......... + \operatorname{Sin} (a + (n - 1)d) = \dfrac{{(\operatorname{Sin} (a + (n - 1)\dfrac{d}{2})\operatorname{Sin} (\dfrac{{nd}}{2})}}{{\operatorname{Sin} \left( {\dfrac{d}{2}} \right)}}} \right]$

By comparing equation (1) with the above formula, we get

\[

a = {10^0},d = {10^0} \\

n - 1 = 35 \\

n = 36 \\

\]

Using the values of a, d, n and substituting these values in the above formula, we get

$

\Rightarrow \dfrac{{\operatorname{Sin} (10 + (36 - 1)\dfrac{{10}}{2})\operatorname{Sin} (\dfrac{{36 \times 10}}{2})}}{{\operatorname{Sin} (\dfrac{{10}}{2})}} \\

\Rightarrow \dfrac{{\operatorname{Sin} (10 + 35 \times 5)\operatorname{Sin} (36 \times 5)}}{{\operatorname{Sin} 5}} \\

$

After simplifying further

$ \Rightarrow \dfrac{{\operatorname{Sin} ({{10}^0} + {{175}^0})\operatorname{Sin} ({{180}^0})}}{{\operatorname{Sin} {5^0}}}$

Since, the value of $\operatorname{Sin} {180^0} = 0$

$

\Rightarrow \dfrac{{\operatorname{Sin} ({{185}^0}) \times 0}}{{\operatorname{Sin} {5^0}}} \\

\Rightarrow 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.

Recently Updated Pages

Calculate the entropy change involved in the conversion class 11 chemistry JEE_Main

The law formulated by Dr Nernst is A First law of thermodynamics class 11 chemistry JEE_Main

For the reaction at rm0rm0rmC and normal pressure A class 11 chemistry JEE_Main

An engine operating between rm15rm0rm0rmCand rm2rm5rm0rmC class 11 chemistry JEE_Main

For the reaction rm2Clg to rmCrmlrm2rmg the signs of class 11 chemistry JEE_Main

The enthalpy change for the transition of liquid water class 11 chemistry JEE_Main

Trending doubts

Difference Between Plant Cell and Animal Cell

Write an application to the principal requesting five class 10 english CBSE

Ray optics is valid when characteristic dimensions class 12 physics CBSE

Give 10 examples for herbs , shrubs , climbers , creepers

Write the 6 fundamental rights of India and explain in detail

Write a letter to the principal requesting him to grant class 10 english CBSE

List out three methods of soil conservation

Fill in the blanks A 1 lakh ten thousand B 1 million class 9 maths CBSE

Write a letter to the Principal of your school to plead class 10 english CBSE