Question Number 76265 by A8;15: last updated on 25/Dec/19
Answered by Tanmay chaudhury last updated on 26/Dec/19
![S=cosα+cos2α+cos3α+...+cosnα multiply each term by 2sin((α/2)) and apply trigo formula and adding them 2cosαsin((α/2))=sin(((3α)/2))−sin((α/2)) 2cos2αsin((α/2))=sin(((5α)/2))−sin(((3α)/2)) ... ... 2cosnαsin((α/2))=sin(((2n+1)/2))α−sin(((2n−1)/2))α add them 2sin((α/2))×S=sin(((2n+1)/2))α−sin((α/2)) S=(1/(2sin((α/2))))×[sin(((2n+1)/2))α−sin((α/2))]](https://www.tinkutara.com/question/Q76300.png)
$${S}={cos}\alpha+{cos}\mathrm{2}\alpha+{cos}\mathrm{3}\alpha+…+{cosn}\alpha \\ $$$${multiply}\:{each}\:{term}\:{by}\:\mathrm{2}{sin}\left(\frac{\alpha}{\mathrm{2}}\right)\:{and}\:{apply}\:{trigo} \\ $$$${formula}\:{and}\:{adding}\:{them} \\ $$$$\mathrm{2}{cos}\alpha{sin}\left(\frac{\alpha}{\mathrm{2}}\right)={sin}\left(\frac{\mathrm{3}\alpha}{\mathrm{2}}\right)−{sin}\left(\frac{\alpha}{\mathrm{2}}\right) \\ $$$$\mathrm{2}{cos}\mathrm{2}\alpha{sin}\left(\frac{\alpha}{\mathrm{2}}\right)={sin}\left(\frac{\mathrm{5}\alpha}{\mathrm{2}}\right)−{sin}\left(\frac{\mathrm{3}\alpha}{\mathrm{2}}\right) \\ $$$$… \\ $$$$… \\ $$$$\mathrm{2}{cosn}\alpha{sin}\left(\frac{\alpha}{\mathrm{2}}\right)={sin}\left(\frac{\mathrm{2}{n}+\mathrm{1}}{\mathrm{2}}\right)\alpha−{sin}\left(\frac{\mathrm{2}{n}−\mathrm{1}}{\mathrm{2}}\right)\alpha \\ $$$$\boldsymbol{{add}}\:\boldsymbol{{them}} \\ $$$$\mathrm{2}{sin}\left(\frac{\alpha}{\mathrm{2}}\right)×\boldsymbol{{S}}={sin}\left(\frac{\mathrm{2}{n}+\mathrm{1}}{\mathrm{2}}\right)\alpha−{sin}\left(\frac{\alpha}{\mathrm{2}}\right) \\ $$$$\boldsymbol{{S}}=\frac{\mathrm{1}}{\mathrm{2}{sin}\left(\frac{\alpha}{\mathrm{2}}\right)}×\left[{sin}\left(\frac{\mathrm{2}{n}+\mathrm{1}}{\mathrm{2}}\right)\alpha−{sin}\left(\frac{\alpha}{\mathrm{2}}\right)\right] \\ $$$$ \\ $$