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Question Number 117606 by TANMAY PANACEA last updated on 12/Oct/20

$$find$$$$sina+sin(a+b)+sin(a+2b)++..+sin\{a+(n-1)b\}$$

Answered by Dwaipayan Shikari last updated on 12/Oct/20

$$\frac{1}{2sin\frac{b}{2}}(2sinasin\frac{b}{2}+2sin(a+b)sin\frac{b}{2}+.....2sin(a+(n-1)b)sin\frac{b}{2})$$$$\frac{1}{2sin\frac{b}{2}}(cos(a-\frac{b}{2})-cos(a+\frac{b}{2})+cos(a+\frac{b}{2})-cos(a+\frac{3b}{2})+........cos(a+nb-\frac{3b}{2})-cos(a+nb-\frac{b}{2}))$$$$=\frac{1}{2sin\frac{b}{2}}(cos(a-\frac{b}{2})-cos(a+nb-\frac{b}{2}))$$$$=\frac{1}{sin\frac{b}{2}}sin(a+\frac{nb}{2}-\frac{b}{2})sin\left(\frac{nb}{2}\right)$$

Commented by TANMAY PANACEA last updated on 12/Oct/20

$$bahdarun...$$

Commented by Dwaipayan Shikari last updated on 12/Oct/20

ধন্যবাদ

Commented by Dwaipayan Shikari last updated on 12/Oct/20

$$Ifb\to 0$$$$\frac{1}{sin\frac{b}{2}}sin(a+\frac{nb}{2}-\frac{b}{2})sin\left(\frac{nb}{2}\right)=\frac{nb}{2}.\frac{1}{\frac{b}{2}}sin\left(a\right)=nsina$$$$Whichistrue$$

Answered by mathmax by abdo last updated on 12/Oct/20

$${\mathrm{A}}_{\mathrm{n}}=\sum _{\mathrm{k}=0}^{\mathrm{n}-1}\sin (\mathrm{a}+\mathrm{kb})\Rightarrow {\mathrm{A}}_{\mathrm{n}}=\mathrm{Im}\left(\sum _{\mathrm{k}=0}^{\mathrm{n}-1}{\mathrm{e}}^{\mathrm{i}(\mathrm{a}+\mathrm{kb})}\right)$$$$\sum _{\mathrm{k}=0}^{\mathrm{n}-1}{\mathrm{e}}^{\mathrm{i}(\mathrm{a}+\mathrm{kb})}={\mathrm{e}}^{\mathrm{ia}}\sum _{\mathrm{k}=0}^{\mathrm{n}-1}{\left({\mathrm{e}}^{\mathrm{ib}}\right)}^{\mathrm{k}}={\mathrm{e}}^{\mathrm{ia}}\times \frac{1-{\left({\mathrm{e}}^{\mathrm{ib}}\right)}^{\mathrm{n}}}{1-{\mathrm{e}}^{\mathrm{ib}}}$$$$={\mathrm{e}}^{\mathrm{ia}}\times \frac{1-{\mathrm{e}}^{\mathrm{inb}}}{1-\mathrm{cosb}-\mathrm{isinb}}={\mathrm{e}}^{\mathrm{ia}}\times \frac{1-\cos \left(\mathrm{nb}\right)-\mathrm{isin}\left(\mathrm{nb}\right)}{1-\mathrm{cosb}-\mathrm{isinb}}$$$$={\mathrm{e}}^{\mathrm{ia}}.\frac{{2\sin }^2\left(\frac{\mathrm{nb}}{2}\right)-2\mathrm{isin}\left(\frac{\mathrm{nb}}{2}\right)\cos \left(\frac{\mathrm{nb}}{2}\right)}{{2\sin }^2\left(\frac{\mathrm{b}}{2}\right)-2\mathrm{isin}\left(\frac{\mathrm{b}}{2}\right)\cos \left(\frac{\mathrm{b}}{2}\right)}$$$$={\mathrm{e}}^{\mathrm{ia}}\times \frac{-\mathrm{isin}\left(\frac{\mathrm{nb}}{2}\right){\mathrm{e}}^{\mathrm{i}\frac{\mathrm{nb}}{2}}}{-\mathrm{isin}\left(\frac{\mathrm{b}}{2}\right){\mathrm{e}}^{\frac{\mathrm{ib}}{2}}}=\frac{\sin \left(\frac{\mathrm{nb}}{2}\right)}{\sin \left(\frac{\mathrm{b}}{2}\right)}{\mathrm{e}}^{\mathrm{ia}}\times {\mathrm{e}}^{\mathrm{i}\frac{(\mathrm{n}-1)\mathrm{b}}{2}}$$$$=\frac{\sin \left(\frac{\mathrm{nb}}{2}\right)}{\sin \left(\frac{\mathrm{b}}{2}\right)}\times {\mathrm{e}}^{\mathrm{i}(\mathrm{a}+\frac{(\mathrm{n}-1)\mathrm{b}}{2})}\Rightarrow {\mathrm{A}}_{\mathrm{n}}=\frac{\sin \left(\frac{\mathrm{nb}}{2}\right)}{\sin \left(\frac{\mathrm{b}}{2}\right)}\times \sin (\mathrm{a}+\frac{(\mathrm{n}-1)\mathrm{b}}{2})$$

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