The value of $\sum_\limits{r=0}^{n}$ n C r (sin rx) is equal to

$\sum_\limits{r=0}^{n}$ n C r sin rx = Im ( $\sum_\limits{r=0}^{n}$ n C r e irx ) = Im ( $\sum_\limits{r=0}^{n}$ n C r (e ix ) r ) = Im ((1 + e ix ) n ) = Im (1 + cosx + i sinx) n = Im (2 cos 2 $\frac{x}{2}$ + 2i sin $\frac{x}{2}$ $\cdot$ cos $\frac{x}{2}$ ) n = Im (2 cos $\frac{x}{2}$ (cos $\frac{x}{2}$ + i sin $\frac{x}{2}$ )) n = 2 n $\cdot$ cos n $\frac{x}{2}$ $\cdot$ sin $\frac{n x}{2}$

Binomial Theorem

Grade 11

Question:

The value of $\sum_\limits{r=0}^{n}$ nCr (sin rx) is equal to

2n $\cdot$ cosn $\frac{x}{2}$ $\cdot$ sin$\frac{n x}{2}$

2n+1 $\cdot$ cosn $\frac{x}{2}$ $\cdot$ sin$\frac{n x}{2}$

2n+1 $\cdot$ sinn $\frac{x}{2}$ $\cdot$ cos$\frac{n x}{2}$

2n $\cdot$ sinn $\frac{x}{2}$ $\cdot$ cos$\frac{n x}{2}$

Step-by-Step Solution

Key Concept: Use Euler's identity to convert the trigonometric sum into the imaginary part of a binomial expansion of (1 + e^{ix})^n.

$\sum_\limits{r=0}^{n}$ nCr sin rx = Im ($\sum_\limits{r=0}^{n}$ nCr eirx) = Im ($\sum_\limits{r=0}^{n}$ nCr (eix)r) = Im ((1 + eix)n) = Im (1 + cosx + i sinx)n = Im (2 cos2 $\frac{x}{2}$ + 2i sin $\frac{x}{2}$ $\cdot$ cos $\frac{x}{2}$)n = Im (2 cos $\frac{x}{2}$ (cos $\frac{x}{2}$ + i sin $\frac{x}{2}$))n = 2n $\cdot$ cosn $\frac{x}{2}$ $\cdot$ sin$\frac{n x}{2}$

Correct Answer: A

Question:

Step-by-Step Solution

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