Question

S = ${\sum }_{r=0}^{10}co{s}^3\frac{r\pi }{3}$ find the value of 16S

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Answer 1

S = ${\sum }_{r=0}^{10}co{s}^3\frac{r\pi }{3}$

Power of each term in the given expression is 3.We are familiar with cosine series and sine series, which has the power of cosine and sine 1. We will try to convert this into terms which have powers one.

We know, cos 3x = 4 $co{s}^{3}x$ - 3 cosx

$co{s}^{3}x$ = $\frac{cos3x+3cosx}{4}$

${\sum }_{r=0}^{10}co{s}^3\frac{r\pi }{3}$ = $\frac{1}{4}{\sum }_{r=0}^{10}[\left(cos3.\frac{r\pi }{3}\right)+3.cos\frac{r\pi }{3}]$

= $\frac{1}{4}{\sum }_{r=0}^{10}[cosr\pi +3.cos\frac{r\pi }{3}]$

= $\frac{1}{4}$ $\left[\right(cos0+cos\pi +cos2\pi +cos3\pi +.............cos10\pi )+3(cos0+cos\frac{\pi }{3}+cos\frac{2\pi }{3}+cos\frac{3\pi }{3}cos\frac{4\pi }{3}+..........11terms)]$

For cosine series $\alpha$ = 0 ,$\beta$ = $\frac{\pi }{3}$ n = 11

= $\frac{1}{4}\left[\right(1-1+1-1+....+1)+3\frac{sin\left(\frac{11.\pi }{2\times 3}\right)}{sin\left(\frac{\pi }{2\times 3}\right)}.cos(0+\frac{(11-1)}{2}.\frac{\pi }{3})]$

= $\frac{1}{4}[1+3\frac{sin\frac{11.\pi }{6}}{sin\frac{\pi }{6}}.cos\frac{5\pi }{3}]$

= $\frac{1}{4}[1+3\frac{sin\left(2\pi \frac{11.\pi }{6}\right)}{sin\frac{\pi }{6}}\times cos(2\pi -\frac{\pi }{3})]$

= $\frac{1}{4}[1+3\frac{-sin\frac{\pi }{6}}{sin\frac{\pi }{6}}\times cos\left(\frac{\pi }{3}\right)]$

= $\frac{1}{4}[1+3(-1)\times \frac{1}{2}]$

= $\frac{1}{4}[1-\frac{3}{2}]$

S = $\frac{1}{4}\times \left(\frac{-1}{2}\right)$ = $-\frac{1}{8}$

$16S=16\times \left(\frac{-1}{8}\right)=-2$