
Find $ \sin \theta + \sin (\pi + \theta ) + \sin (2\pi + \theta ) + \sin (3\pi + \theta ) + $ ..... upto $ 2021 $ terms.
Answer
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Hint: The trigonometric functions are the circular functions with the function of an angle and are related to the angles of a triangle with the lengths of its sides. In the Cartesian coordinate system, the circle centred origin $ o(0,0) $ is the unit circle, where the points distance from the origin is always one. The Y-coordinate to the point of intersection is equal to the $ \sin \theta $ . Here, we will use the properties of the cosine functions and the All STC rule.
Complete step-by-step answer:
By using the All STC rule, sine function is positive in the first and second quadrant and sine function is in the third and fourth quadrant.
The given series is in the combination of odd and even angles.
Here, we will write one equivalent relation for both the odd combination angles and even combination angles.
\[\Rightarrow \sin (\pi + \theta ) = \sin (3\pi + \theta ) = \sin [(2n + 1)\pi + \theta ] = - \sin \theta \] ..... (a)
Similarly, for the even functions –
\[\Rightarrow \sin (2\pi + \theta ) = \sin (4\pi + \theta ) = \sin [2n\pi + \theta ] = \sin \theta \] ...... (b)
By using the values of the equation (a) and the equation (b)
Now, adding the two consecutive terms –
First two terms-
$ \sin \theta + \sin (\pi + \theta ) = 0 $
Adding next two terms –
$\Rightarrow \sin (2\pi + \theta ) + \sin (3\pi + \theta ) = 0 $
.....
Similarly continue adding all the terms till $ 2021 $ terms –
$\Rightarrow \sin (2018\pi + \theta ) + \sin (2019\pi + \theta ) = 0 $
$\Rightarrow \sin (2020\pi + \theta ) + \sin (2021\pi + \theta ) = 0 $
From all the above results - sum of all the terms gives us the resultant value as the zero.
Hence, the required answer - $ \sin \theta + \sin (\pi + \theta ) + \sin (2\pi + \theta ) + \sin (3\pi + \theta ) + $ ..... upto $ 2021 $ is $ 0 $
Note: Remember the All STC rule, it is also known as the ASTC rule in geometry. It states that all the trigonometric ratios in the first quadrant ( $ 0^\circ \;{\text{to 90}}^\circ $ ) are positive, sine and cosec are positive in the second quadrant ( $ 90^\circ {\text{ to 180}}^\circ $ ), tan and cot are positive in the third quadrant ( $ 180^\circ \;{\text{to 270}}^\circ $ ) and sin and cosec are positive in the fourth quadrant ( $ 270^\circ {\text{ to 360}}^\circ $ ).
Complete step-by-step answer:
By using the All STC rule, sine function is positive in the first and second quadrant and sine function is in the third and fourth quadrant.
The given series is in the combination of odd and even angles.
Here, we will write one equivalent relation for both the odd combination angles and even combination angles.
\[\Rightarrow \sin (\pi + \theta ) = \sin (3\pi + \theta ) = \sin [(2n + 1)\pi + \theta ] = - \sin \theta \] ..... (a)
Similarly, for the even functions –
\[\Rightarrow \sin (2\pi + \theta ) = \sin (4\pi + \theta ) = \sin [2n\pi + \theta ] = \sin \theta \] ...... (b)
By using the values of the equation (a) and the equation (b)
Now, adding the two consecutive terms –
First two terms-
$ \sin \theta + \sin (\pi + \theta ) = 0 $
Adding next two terms –
$\Rightarrow \sin (2\pi + \theta ) + \sin (3\pi + \theta ) = 0 $
.....
Similarly continue adding all the terms till $ 2021 $ terms –
$\Rightarrow \sin (2018\pi + \theta ) + \sin (2019\pi + \theta ) = 0 $
$\Rightarrow \sin (2020\pi + \theta ) + \sin (2021\pi + \theta ) = 0 $
From all the above results - sum of all the terms gives us the resultant value as the zero.
Hence, the required answer - $ \sin \theta + \sin (\pi + \theta ) + \sin (2\pi + \theta ) + \sin (3\pi + \theta ) + $ ..... upto $ 2021 $ is $ 0 $
Note: Remember the All STC rule, it is also known as the ASTC rule in geometry. It states that all the trigonometric ratios in the first quadrant ( $ 0^\circ \;{\text{to 90}}^\circ $ ) are positive, sine and cosec are positive in the second quadrant ( $ 90^\circ {\text{ to 180}}^\circ $ ), tan and cot are positive in the third quadrant ( $ 180^\circ \;{\text{to 270}}^\circ $ ) and sin and cosec are positive in the fourth quadrant ( $ 270^\circ {\text{ to 360}}^\circ $ ).
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