Find the solution of the equation given below
$\operatorname{Sin} {10^0} + \operatorname{Sin} {20^0} + \operatorname{Sin} {30^0} + ............ + \operatorname{Sin} {360^0}$
Answer
384.9k+ 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.
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