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Question Number 52841 by Tawa1 last updated on 13/Jan/19

Commented by maxmathsup by imad last updated on 14/Jan/19

$$\pi \to {180}^o$$$$\alpha \to k^o\Rightarrow {180}^{}.\alpha =k\pi \Rightarrow \alpha =\frac{k\pi }{180}\Rightarrow$$$$S=\sum _{k=0}^{90}{sin}^2\left(\frac{k\pi }{180}\right)generallyletdetermine{S}_n=\sum _{k=0}^n{sin}^2\left(\frac{k\pi }{180}\right)\Rightarrow$$$$S_n=\sum _{k=0}^n\frac{1-cos\left(\frac{k\pi }{90}\right)}{2}=\frac{n+1}{2}-\frac{1}{2}\sum _{k=0}^{n}cos\left(\frac{k\pi }{90}\right)but$$$$\sum _{k=0}^{n}cos\left(\frac{k\pi }{90}\right)=Re\left(\sum _{k=0}^n{e}^{i\frac{k\pi }{90}}\right)and\sum _{k=0}^n{e}^{i\frac{k\pi }{90}}=\sum _{k=0}^n{\left(e^{\frac{i\pi }{90}}\right)}^k$$$$=\frac{1-e^{i(n+1)\frac{\pi }{90}}}{1-e^{\frac{i\pi }{90}}}=\frac{1-cos\left(\frac{(n+1)\pi }{90}\right)-isin\left(\frac{(n+1)\pi }{90}\right)}{1-cos\left(\frac{\pi }{90}\right)-isin\left(\frac{\pi }{90}\right)}$$$$=\frac{2{sin}^2\left(\frac{(n+1)\pi }{180}\right)-2isin\left(\frac{(n+1)\pi }{180}\right)cos\left(\frac{(n+1)\pi }{180}\right)}{2{sin}^2\left(\frac{\pi }{180}\right)-2isin\left(\frac{\pi }{180}\right)cos\left(\frac{\pi }{180}\right)}$$$$=\frac{-isin\left(\frac{(n+1)\pi }{180}\right)\left(e^{i\frac{(n+1)\pi }{180}}\right)}{-isin\left(\frac{\pi }{180}\right)e^{\frac{i\pi }{180}}}=\frac{sin\left(\frac{(n+1)\pi }{180}\right)}{sin\left(\frac{\pi }{180}\right)}{e}^{\frac{in\pi }{180}}\Rightarrow$$$$\sum _{k=0}^{n}cos\left(\frac{k\pi }{90}\right)=\frac{sin\left(\frac{(n+1)\pi }{180}\right)cos\left(\frac{n\pi }{180}\right)}{sin\left(\frac{\pi }{180}\right)}\Rightarrow$$$$S_n=\frac{n+1}{2}-\frac{sin\left(\frac{(n+1)\pi }{180}\right)cos\left(\frac{n\pi }{180}\right)}{sin\left(\frac{\pi }{180}\right)}\Rightarrow S=S_{90}=\frac{91}{2}-\frac{sin\left(\frac{91\pi }{180}\right)cos\left(\frac{\pi }{2}\right)}{sin\left(\frac{\pi }{180}\right)}$$$$=\frac{91}{2}-0\Rightarrow S=\frac{91}{2}.$$$$$$

Commented by Tawa1 last updated on 16/Jan/19

$$\mathrm{God}\mathrm{bless}\mathrm{you}\mathrm{sir}$$

Commented by Tawa1 last updated on 17/Jan/19

$$\mathrm{Sir},\mathrm{please}\mathrm{explain}\mathrm{from}\mathrm{here}\mathrm{sir}.$$$$\mathrm{How}{\mathrm{S}}_{\mathrm{n}}=\sum _{\mathrm{k}=0}^{\mathrm{n}}{\sin }^2\left(\frac{\mathrm{k}\pi }{180}\right)\mathrm{becomes}$$$${\mathrm{S}}_{\mathrm{n}}=\sum _{\mathrm{k}=0}^{\mathrm{n}}\frac{1-\cos \left(\frac{\mathrm{k}\pi }{90}\right)}{2}=\frac{\mathrm{n}+1}{2}-\frac{1}{2}\sum _{\mathrm{k}=0}^{\mathrm{n}}\cos \left(\frac{\mathrm{k}\pi }{90}\right)$$$$$$$$\mathrm{And}\mathrm{how}2\mathrm{becomes}1-{\mathrm{e}}^{\mathrm{i}\frac{\pi }{90}}$$$$\mathrm{please}\mathrm{sir}.$$

Commented by maxmathsup by imad last updated on 19/Jan/19

$$sirtakex=\frac{k\pi }{180}anduseformulalae{sin}^2\left(x\right)=\frac{1-cos\left(2x\right)}{2}$$$$also\sum _{k=0}^n\frac{1-cos\left(\frac{k\pi }{90}\right)}{2}=\frac{1}{2}\sum _{k=0}^n\left(1\right)-\frac{1}{2}\sum _{k=0}^{n}cos\left(\frac{k\pi }{90}\right)and$$$$\sum _{k=0}^n\left(1\right)=n+1(lookthat\sum _{k=0}^{n}a=(n+1)a)$$

Answered by tanmay.chaudhury50@gmail.com last updated on 14/Jan/19

$$1+89=90so{sin}^{2}1+{sin}^{2}89$$$$={sin}^{2}1+{sin}^2(90-1)$$$$={sin}^{2}1+{cos}^{2}1$$$$=1$$$$similarly..$$$$2+88=90$$$$...$$$$...$$$$45+45=90$$$$$$$$s=({sin}^{2}1+{sin}^{2}2+{sin}^{2}3+...+{sin}^{2}89)+{sin}^{2}90$$$$s=({sin}^{2}89+{sin}^{2}88+{sin}^{2}87+..+{sin}^{2}1)+{sin}^{2}90$$$$2s=[({sin}^{2}1+{sin}^{2}89)+({sin}^{2}2+{sin}^{2}88)+..+({sin}^{2}89+{sin}^{2}1)]+(1+1)$$$$2s=[1+1+1...89times]+2$$$$s=\frac{91}{2}=45.5$$$$$$

Commented by tanmay.chaudhury50@gmail.com last updated on 14/Jan/19

$$thankyou..Godblessall...$$

Commented by Tawa1 last updated on 14/Jan/19

$$\mathrm{God}\mathrm{bless}\mathrm{you}\mathrm{sir}$$

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