Question

$\sf{cosA+cos2A+cos3A+ \cdot \cdot \cdot \cdot \cdot \cdot \:cosnA}$​

Answer 1

Step-by-step explanation:

We have,

$\tt{ cos(A) + cos(2A) + cos(3A) + \cdots + cos(nA) }$

$\displaystyle \tt{ = \sum^{n}_{k = 1} cos(kA) }$

$\displaystyle \tt{ = \sum^{n}_{k = 1} \dfrac{ {e}^{i(kA)} +{e}^{ - i(kA)} }{2} }$

$\displaystyle \tt{ = \sum^{n}_{k = 1} \dfrac{ {e}^{i(kA)} }{2} + \sum^{n}_{k = 1} \dfrac{ {e}^{ - i(kA)} }{2} }$

$\displaystyle \tt{ = \dfrac{1}{2} \sum^{n}_{k = 1} {e}^{i(kA)} + \dfrac{1}{2} \sum^{n}_{k = 1} {e}^{ - i(kA)} }$

$\displaystyle \tt{ = \dfrac{1}{2} \left \{ {e}^{i(A)} + {e}^{i(2A)} + {e}^{i(3A)} + \cdots +{e}^{i(nA)} \right \} + \dfrac{1}{2} \left \{ {e}^{ - i(A)} + {e}^{ - i(2A)} + {e}^{ - i(3A)} + \cdots +{e}^{ - i(nA)} \right \} }$

$\displaystyle \tt{ = \dfrac{1}{2} \left \{ {e}^{iA} + \left({e}^{iA} \right)^{2} + \left({e}^{iA} \right)^{3} + \cdots + \left({e}^{iA} \right)^{n} \right \} + \dfrac{1}{2} \left \{ {e}^{ - iA} + \left({e}^{ - iA} \right)^{2} + \left({e}^{ - iA} \right)^{3} + \cdots + \left({e}^{ - iA} \right)^{n} \right \} }$

$\displaystyle \tt{ = \dfrac{ {e}^{iA} }{2} \left \{ 1 + \left({e}^{iA} \right) + \left({e}^{iA} \right)^{2} + \cdots + \left({e}^{iA} \right)^{n - 1} \right \} + \dfrac{{e}^{ - iA}}{2} \left \{ 1 + \left({e}^{ - iA} \right) + \left({e}^{ - iA} \right)^{2} + \cdots + \left({e}^{ - iA} \right)^{n - 1} \right \} }$

$\displaystyle \tt{ = \dfrac{ {e}^{iA} }{2} \left \{ \dfrac{ \left({e}^{iA} \right)^{n} - 1}{{e}^{iA} - 1}\right \} + \dfrac{{e}^{ - iA}}{2} \left \{ \dfrac{ \left({e}^{ - iA} \right)^{n} - 1}{{e}^{ - iA} - 1} \right \} }$

$\displaystyle \tt{ = \dfrac{ {e}^{iA} }{2} \left \{ \dfrac{ {e}^{i(nA)} - 1}{{e}^{iA} - 1}\right \} + \dfrac{{e}^{ - iA}}{2} \left \{ \dfrac{ {e}^{ - i(nA)} - 1}{{e}^{ - iA} - 1} \right \} }$

$\displaystyle \tt{ = \dfrac{ {e}^{iA} }{2} \left \{ \dfrac{ {e}^{i( \frac{n}{2})A} - {e}^{ - i( \frac{n}{2})A}}{{e}^{i (\frac{A}{2})} -{e}^{ - i (\frac{A}{2})}}\right \} \cdot \dfrac{ {e}^{i( \frac{n}{2})A}}{ {e}^{i( \frac{A}{2})}} + \dfrac{{e}^{ - iA}}{2} \left \{ \dfrac{ {e}^{ - i( \frac{n}{2})A} - {e}^{ i( \frac{n}{2})A}}{{e}^{ - i (\frac{A}{2})} -{e}^{ i (\frac{A}{2})}}\right \} \cdot \dfrac{ {e}^{ - i( \frac{n}{2})A}}{ {e}^{ - i( \frac{A}{2})}} }$

$\displaystyle \tt{ = \dfrac{ 1 }{2} \left \{ \dfrac{ {e}^{i( \frac{n}{2})A} - {e}^{ - i( \frac{n}{2})A}}{{e}^{i (\frac{A}{2})} -{e}^{ - i (\frac{A}{2})}}\right \} \cdot {e}^{i( \frac{n}{2})A} \cdot{e}^{i( \frac{A}{2})} + \dfrac{1}{2} \left \{ \dfrac{ {e}^{ - i( \frac{n}{2})A} - {e}^{ i( \frac{n}{2})A}}{{e}^{ - i (\frac{A}{2})} -{e}^{ i (\frac{A}{2})}}\right \} \cdot {e}^{ - i( \frac{n}{2})A} \cdot{e}^{ - i( \frac{A}{2})} }$

$\displaystyle \tt{ = \dfrac{ 1 }{2} \left \{ \dfrac{ {e}^{i( \frac{n}{2})A} - {e}^{ - i( \frac{n}{2})A}}{{e}^{i (\frac{A}{2})} -{e}^{ - i (\frac{A}{2})}}\right \} \cdot {e}^{i( \frac{n + 1}{2})A} + \dfrac{1}{2} \left \{ \dfrac{ {e}^{ - i( \frac{n}{2})A} - {e}^{ i( \frac{n}{2})A}}{{e}^{ - i (\frac{A}{2})} -{e}^{ i (\frac{A}{2})}}\right \} \cdot {e}^{ - i( \frac{n + 1}{2})A} }$

$\displaystyle \tt{ = \dfrac{ {e}^{ i( \frac{n + 1}{2})A}+ {e}^{ - i( \frac{n + 1}{2})A}}{2} \left \{ \dfrac{ {e}^{i( \frac{n}{2})A} - {e}^{ - i( \frac{n}{2})A}}{{e}^{i (\frac{A}{2})} -{e}^{ - i (\frac{A}{2})}}\right \} }$

$\displaystyle \tt{ = cos \left( \dfrac{(n + 1)A}{2} \right) \left \{ \dfrac{ sin \left(\dfrac{nA}{2} \right) }{sin \left(\dfrac{A}{2} \right)}\right \} }$

$\displaystyle \tt{ = cos \left \{\dfrac{(n + 1)A}{2} \right \} \cdot \dfrac{ sin \left(\dfrac{nA}{2} \right) }{sin \left(\dfrac{A}{2} \right)}}$

Answer 2

$\large\underline{\sf{Solution-}}$

Given expression is

$\rm :\longmapsto\: \sf{cosA+cos2A+cos3A+ \cdot \cdot \cdot \cdot \cdot \cdot \:cosnA}$

Before we derive the sum of above series, Let we first derive an expression for

$\rm :\longmapsto\: \: cosa + cos(a + b) + cos(a + 2b) + - - - + cos[a + (n - 1)b]$

can be rewritten as

$\rm \:  =  \: \displaystyle\sum_{k=0}^{n-1}\sf cos(a + kb)$

can be further rewritten as

$\rm \:  =  \:\dfrac{1}{2sin\dfrac{b}{2} } \displaystyle\sum_{k=0}^{n-1}\sf 2 \: cos(a + kb) \: sin\dfrac{b}{2}$

We know,

$\boxed{\tt{ 2sinx \: cosy \: = \: sin(x + y) + sin(x - y) \: }}$

So, using this identity we get

$\rm \:  =  \:\dfrac{1}{2sin\dfrac{b}{2} } \displaystyle\sum_{k=0}^{n-1}\sf \bigg(sin\bigg[\dfrac{b}{2} + a + kb\bigg] + sin\bigg[\dfrac{b}{2} - a - kb \bigg]\bigg)$

$\rm \:  =  \:\dfrac{1}{2sin\dfrac{b}{2} } \displaystyle\sum_{k=0}^{n-1}\sf \bigg(sin\bigg[\dfrac{2a + (2k + 1)b}{2}\bigg] - sin\bigg[\dfrac{2a + (2k - 1)b}{2}\bigg]\bigg)$

$\rm \:  =  \:\dfrac{1}{2sin\dfrac{b}{2} } \displaystyle\sum_{k=0}^{n-1}\sf \bigg(sin\bigg[a + \dfrac{(2k + 1)b}{2}\bigg] - sin\bigg[a + \dfrac{(2k - 1)b}{2}\bigg]\bigg)$

$\rm \:  =  \:\dfrac{1}{2sin\dfrac{b}{2} } \sf \bigg(sin\bigg[a + \dfrac{(2n - 1)b}{2}\bigg] - sin\bigg[a - \dfrac{b}{2} \bigg]\bigg)$

We know,

$\boxed{\tt{ sinx - siny = 2cos\bigg[\dfrac{x + y}{2} \bigg] \: sin\bigg[\dfrac{x - y}{2} \bigg]}}$

$\rm \:  =  \:\dfrac{1}{2sin\dfrac{b}{2} } \sf \bigg(2cos \dfrac{1}{2} \bigg[a + \dfrac{(2n - 1)b}{2} + a - \dfrac{b}{2} \bigg]sin \dfrac{1}{2} \bigg[a + \dfrac{(2n - 1)b}{2} - a + \dfrac{b}{2}\bigg]\bigg)$

$\rm \:  =  \: \dfrac{cos\bigg[a + \dfrac{(n - 1)b}{2} \bigg]sin\bigg[\dfrac{nb}{2} \bigg]}{sin\bigg[\dfrac{b}{2} \bigg]}$

So, we get

$\rm \: \: cosa + cos(a + b) + cos(a + 2b) + - - - + cos[a + (n - 1)b] \\ \\ \rm \:  =  \: \dfrac{cos\bigg[a + \dfrac{(n - 1)b}{2} \bigg]sin\bigg[\dfrac{nb}{2} \bigg]}{sin\bigg[\dfrac{b}{2} \bigg]} \: \: \: \: \: \: \: \:$

Now, we replace a by A and b by A, we get

$\rm :\longmapsto\:\sf{cosA+cos2A+cos3A+ \cdot \cdot \cdot \cdot \cdot \cdot \:cosnA}$

$\rm \:  =  \: \dfrac{cos\bigg[A + \dfrac{(n - 1)A}{2} \bigg]sin\bigg[\dfrac{nA}{2} \bigg]}{sin\bigg[\dfrac{A}{2} \bigg]}$

$\rm \:  =  \: \dfrac{cos\bigg[ \dfrac{2A + (n - 1)A}{2} \bigg]sin\bigg[\dfrac{nA}{2} \bigg]}{sin\bigg[\dfrac{A}{2} \bigg]}$

$\rm \:  =  \: \dfrac{cos\bigg[ \dfrac{2A + nA - A}{2} \bigg]sin\bigg[\dfrac{nA}{2} \bigg]}{sin\bigg[\dfrac{A}{2} \bigg]}$

$\rm \:  =  \: \dfrac{cos\bigg[ \dfrac{A + nA}{2} \bigg]sin\bigg[\dfrac{nA}{2} \bigg]}{sin\bigg[\dfrac{A}{2} \bigg]}$

$\rm \:  =  \: \dfrac{cos\bigg[ \dfrac{(n + 1)A}{2} \bigg]sin\bigg[\dfrac{nA}{2} \bigg]}{sin\bigg[\dfrac{A}{2} \bigg]}$

Hence,

$\rm :\longmapsto\:\rm{cosA+cos2A+cos3A+ \cdot \cdot \cdot \cdot \cdot \cdot \:cosnA} \\ \\ \rm \:  =  \: \dfrac{cos\bigg[ \dfrac{(n + 1)A}{2} \bigg]sin\bigg[\dfrac{nA}{2} \bigg]}{sin\bigg[\dfrac{A}{2} \bigg]}$