Question and Answers Forum

All Questions      Topic List

Trigonometry Questions

Previous in All Question      Next in All Question      

Previous in Trigonometry      Next in Trigonometry      

Question Number 143439 by ERA last updated on 14/Jun/21

$$\mathrm{sinx}+\sin 2\mathrm{x}+\sin 3\mathrm{x}+.....+\mathrm{sinkx}=\sin \frac{\mathrm{kx}}{2}\times \frac{\sin \frac{\mathrm{k}+1}{2}\mathrm{x}}{\sin \frac{\mathrm{x}}{2}}$$$$\mathrm{prove}$$

Answered by Mathspace last updated on 14/Jun/21

$$A_n=sinx+sin\left(2x\right)+...+sin\left(nx\right)\Rightarrow$$$$A_n=\sum _{k=0}^{n}sin\left(kx\right)=Im\left(\sum _{k=0}^n{e}^{ikx}\right)but$$$$\sum _{k=0}^n{e}^{ikx}=\sum _{k=0}^n{\left(e^{ix}\right)}^k$$$$=\frac{1-{\left(e^{ix}\right)}^{n+1}}{1-e^{ix}}=\frac{1-e^{i(n+1)x}}{1-e^{ix}}$$$$=\frac{1-cos(n+1)x-isin(n+1)x}{1-cosx-isinx}$$$$=\frac{2{sin}^2\left(\frac{(n+1)x}{2}\right)-2isin\left(\frac{(n+1)x}{2}\right)cos\left(\frac{(n+1)x}{2}\right)}{2{sin}^2\left(\frac{x}{2}\right)-2isin\left(\frac{x}{2}\right)cos\left(\frac{x}{2}\right)}$$$$=\frac{-isin\left(\frac{(n+1)x}{2}\right)e^{i\left(\frac{n+1)x}{2}\right)}}{-isin\left(\frac{x}{2}\right)e^{\frac{ix}{2}}}$$$$=\frac{sin\left(\frac{(n+1)x}{2}\right)}{sin\left(\frac{x}{2}\right)}{e}^{\frac{inx}{2}}$$$$\Rightarrow A_n=\frac{sin\left(\frac{(n+1)x}{2}\right)sin\left(\frac{nx}{2}\right)}{sin\left(\frac{x}{2}\right)}$$$$withx\ne 2k\pi$$

Answered by gsk2684 last updated on 14/Jun/21

$$\underset{\mathrm{t}=1}{\sum ^{\mathrm{k}}}\sin \mathrm{tx}=\mathrm{im}\underset{\mathrm{t}=1}{\sum ^{\mathrm{k}}}{\mathrm{e}}^{\mathrm{itx}}$$$$=\mathrm{im}\underset{\mathrm{t}=1}{\sum ^{\mathrm{k}}}{\left({\mathrm{e}}^{\mathrm{ix}}\right)}^{\mathrm{t}}$$$$=\mathrm{im}\frac{{\mathrm{e}}^{\mathrm{ix}}({\left({\mathrm{e}}^{\mathrm{ix}}\right)}^{\mathrm{k}}-1)}{{\mathrm{e}}^{\mathrm{ix}}-1}$$$$=\mathrm{im}\frac{{\mathrm{e}}^{\mathrm{i}(\mathrm{k}+1)\mathrm{x}}-{\mathrm{e}}^{\mathrm{ix}}}{{\mathrm{e}}^{\mathrm{ix}}-1}$$$$=\mathrm{im}\left\{({\mathrm{e}}^{\mathrm{i}(\mathrm{k}+1)\mathrm{x}}-{\mathrm{e}}^{\mathrm{ix}})({\mathrm{e}}^{-\mathrm{ix}}-1)/({(\cos \mathrm{x}-1)}^2+{\sin }^2\mathrm{x})\right\}$$$$=\mathrm{im}\frac{{\mathrm{e}}^{\mathrm{ikx}}-{\mathrm{e}}^{\mathrm{i}(\mathrm{k}+1)\mathrm{x}}-1+{\mathrm{e}}^{\mathrm{ix}}}{2(1-\cos \mathrm{x})}$$$$=(\sin \mathrm{kx}+\sin \mathrm{x}-\sin (\mathrm{k}+1)\mathrm{x})/\left(4{\sin }^2\frac{\mathrm{x}}{2}\right)$$$$=\frac{2\sin \frac{(\mathrm{k}+1)\mathrm{x}}{2}\cos \frac{(\mathrm{k}-1)\mathrm{x}}{2}-2\sin \frac{(\mathrm{k}+1)\mathrm{x}}{2}\cos \frac{(\mathrm{k}+1)\mathrm{x}}{2}}{{4\sin }^2\frac{\mathrm{x}}{2}}$$$$=\{2\sin \frac{(\mathrm{k}+1)\mathrm{x}}{2}.[\cos (\frac{\mathrm{kx}}{2}-\frac{\mathrm{x}}{2})-\cos (\frac{\mathrm{kx}}{2}+\frac{\mathrm{x}}{2})]\}/{4\sin }^2\frac{\mathrm{x}}{2}$$$$=\frac{2\sin \frac{(\mathrm{k}+1)\mathrm{x}}{2}.2\sin \frac{\mathrm{kx}}{2}\sin \frac{\mathrm{x}}{2}}{{4\sin }^2\frac{\mathrm{x}}{2}}$$

Terms of Service

Privacy Policy

Contact: info@tinkutara.com