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# The value of $\sum\limits_{k = 1}^{100} {\sin \left( {kx} \right)\cos \left( {101 - k} \right)x}$ is equal toA. $\dfrac{{101}}{2}\sin \left( {101x} \right)$B. $99\sin \left( {101x} \right)$C. $50\sin \left( {101x} \right)$D. $100\sin \left( {101x} \right)$

Last updated date: 19th Jun 2024
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Hint: Here, we will expand the given expression as the sum of 100 terms, by substituting the value of $k$ from 1 to 100 in the summation. We will find the summation in the reverse order and add both the summation. We will then use the trigonometric formula to solve it further and find the required value of the given expression.

Formula Used:
We will use the formula $\sin \left( {A + B} \right) = \sin A\cos B + \cos A\sin B$.

Let the value of $S = \sum\limits_{k = 1}^{100} {\sin \left( {kx} \right)\cos \left( {101 - k} \right)x}$.
Now, in this question, the sigma symbol , $\sum {}$ denotes the sum of multiple terms starting from the value of $k = 1$ to the value of $k = 100$.
In $S = \sum\limits_{k = 1}^{100} {\sin \left( {kx} \right)\cos \left( {101 - k} \right)x}$, if we substitute the values of $k = 1,2,3,....,100$, then we can write this summation as:
$S = \sin x\cos 100x + \sin 2x\cos 99x + ....\sin 100x\cos x$………………….. $\left( 1 \right)$
Now, rewriting the equation $\left( 1 \right)$ by writing it in the reverse order, we get,
$S = \sin 100x\cos x + \sin 99x\cos 2x + .....\sin x\cos 100x$…………………. $\left( 2 \right)$
Now, adding the equations $\left( 1 \right)$ and $\left( 2 \right)$,
$2S = \left( {\sin x\cos 100x + \sin 100x\cos x} \right) + \left( {\sin 2x\cos 99x + \sin 99x\cos 2x} \right) + ...\left( {\sin 100x\cos x + \sin x\cos 100x} \right)$
Now, using the formula $\sin \left( {A + B} \right) = \sin A\cos B + \cos A\sin B$, we get
$\Rightarrow 2S = \left[ {\sin \left( {x + 100x} \right) + \sin \left( {2x + 99x} \right) + ..\sin \left( {100x + x} \right)} \right]$
Adding the terms in the bracket, we get
$\Rightarrow 2S = \left[ {\sin \left( {101x} \right) + \sin \left( {101x} \right) + ..\sin \left( {101x} \right)} \right]$
As the total number of terms are 100, so we get
$\Rightarrow 2S = 100\sin \left( {101x} \right)$
Dividing both sides by 2, we get
$\Rightarrow S = 50\sin \left( {101x} \right)$
But, we have assumed that $S = \sum\limits_{k = 1}^{100} {\sin \left( {kx} \right)\cos \left( {101 - k} \right)x}$
Therefore, we can say that,
$\sum\limits_{k = 1}^{100} {\sin \left( {kx} \right)\cos \left( {101 - k} \right)x} = 50\sin \left( {101x} \right)$
Hence, the value of $\sum\limits_{k = 1}^{100} {\sin \left( {kx} \right)\cos \left( {101 - k} \right)x}$ is equal to $50\sin \left( {101x} \right)$
Therefore, option C is the correct answer.

Note: A summation means the act of adding or doing a cumulative sum of the given element by substituting the different values of the same variable in the same element and adding them together. Now, the basic difference between a summation and a sigma is that, summation is the adding up of the given series of elements whereas, a sigma is just a mathematical symbol used to indicate this summation without stating anything. Hence, summation plays an important role for finding out the aggregate value of a given element from its lower limit to upper limit of summation.