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# The value of $\sin {1^0} + \sin {2^0} + \sin {3^0} + ...... + \sin {359^0}$ isa. 0b. 1c. -1d. 180

Last updated date: 23rd May 2024
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Hint: Here, we will solve the given equation by checking whether the number of terms is even or not and applying the relevant trigonometric formulae.

$\sin ({360^0} - \theta ) = - \sin \theta \to (1)$
$\Rightarrow \sin {1^0} + \sin {2^0} + \sin {3^0} + .......\sin {359^0} \\ \Rightarrow \sin {1^0} + \sin {2^0} + \sin {3^0} + .......\sin {357^0} + \sin {358^0} + \sin {359^0} \\ \Rightarrow \sin {1^0} + \sin {2^0} + \sin {3^0} + .......\sin ({360^0} - {3^0}) + \sin ({360^0} - {2^0}) + \sin ({360^0} - {1^0}) \\$
$\Rightarrow \sin {1^0} + \sin {2^0} + \sin {3^0} + ....... - \sin {3^0} - \sin {2^0} - \sin {1^0} \to (2)$
Now you know number of terms in series $(1,2,3............359)$ is even
$\Rightarrow 0 + 0 + 0 + 0 + .............. + 0 = 0$