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Question Number 207219 by hardmath last updated on 09/May/24
Find: ((tg 20)/(1 + tg^2 20)) + ((tg 21)/(1 + tg^2 21)) +...+ ((tg 70)/(1 + tg^2 70))
$$\mathrm{Find}: \\ $$$$\frac{\mathrm{tg}\:\mathrm{20}}{\mathrm{1}\:+\:\mathrm{tg}^{\mathrm{2}} \:\mathrm{20}}\:+\:\frac{\mathrm{tg}\:\mathrm{21}}{\mathrm{1}\:+\:\mathrm{tg}^{\mathrm{2}} \:\mathrm{21}}\:+…+\:\frac{\mathrm{tg}\:\mathrm{70}}{\mathrm{1}\:+\:\mathrm{tg}^{\mathrm{2}} \:\mathrm{70}} \\ $$
Answered by Berbere last updated on 10/May/24
((tg(a))/(1+tg^2 (a)))=(((sin(a))/(cos(a)))/(1+((sin^2 (a))/(cos^2 (a)))))=((sin(2a))/2) S=(1/2)Σ_(k=20) ^(70) sin(2k)=(1/2)Σ_(k=20) ^(70) Ime^(2ik) =((Im)/2).e^(i40) .((1−e^(102i) )/(1−e^(2i) ))=(( 1)/2).((.sin(51))/(sin(1)))
$$\frac{{tg}\left({a}\right)}{\mathrm{1}+{tg}^{\mathrm{2}} \left({a}\right)}=\frac{\frac{{sin}\left({a}\right)}{{cos}\left({a}\right)}}{\mathrm{1}+\frac{{sin}^{\mathrm{2}} \left({a}\right)}{{cos}^{\mathrm{2}} \left({a}\right)}}=\frac{{sin}\left(\mathrm{2}{a}\right)}{\mathrm{2}} \\ $$$${S}=\frac{\mathrm{1}}{\mathrm{2}}\underset{{k}=\mathrm{20}} {\overset{\mathrm{70}} {\sum}}{sin}\left(\mathrm{2}{k}\right)=\frac{\mathrm{1}}{\mathrm{2}}\underset{{k}=\mathrm{20}} {\overset{\mathrm{70}} {\sum}}{Ime}^{\mathrm{2}{ik}} \\ $$$$=\frac{{Im}}{\mathrm{2}}.{e}^{{i}\mathrm{40}} .\frac{\mathrm{1}−{e}^{\mathrm{102}{i}} }{\mathrm{1}−{e}^{\mathrm{2}{i}} }=\frac{\:\mathrm{1}}{\mathrm{2}}.\frac{.{sin}\left(\mathrm{51}\right)}{{sin}\left(\mathrm{1}\right)} \\ $$
Commented by Berbere last updated on 10/May/24
yes Σ_(k=1) ^n f(2k)≠Σ_(k=2) ^(2n) f(k);firts sum containe n; seconde 2n−1 thermd
$${yes}\:\underset{{k}=\mathrm{1}} {\overset{{n}} {\sum}}{f}\left(\mathrm{2}{k}\right)\neq\underset{{k}=\mathrm{2}} {\overset{\mathrm{2}{n}} {\sum}}{f}\left({k}\right);{firts}\:{sum}\:{containe} \\ $$$${n};\:{seconde}\:\mathrm{2}{n}−\mathrm{1}\:{thermd} \\ $$
Commented by Frix last updated on 10/May/24
I think this step is wrong: Σ_(k=20) ^(70) sin 2k ≠ Σ_(k=40) ^(140) sin k because the first sum contains 51 items but the second sum contains 101 items
$$\mathrm{I}\:\mathrm{think}\:\mathrm{this}\:\mathrm{step}\:\mathrm{is}\:\mathrm{wrong}: \\ $$$$\underset{{k}=\mathrm{20}} {\overset{\mathrm{70}} {\sum}}\mathrm{sin}\:\mathrm{2}{k}\:\neq\:\underset{{k}=\mathrm{40}} {\overset{\mathrm{140}} {\sum}}\mathrm{sin}\:{k} \\ $$$$\mathrm{because}\:\mathrm{the}\:\mathrm{first}\:\mathrm{sum}\:\mathrm{contains}\:\mathrm{51}\:\mathrm{items} \\ $$$$\mathrm{but}\:\mathrm{the}\:\mathrm{second}\:\mathrm{sum}\:\mathrm{contains}\:\mathrm{101}\:\mathrm{items} \\ $$
Commented by Frix last updated on 10/May/24
My result is still different from yours. I get ((sin 51°)/(2sin 1°))≈22.2647
$$\mathrm{My}\:\mathrm{result}\:\mathrm{is}\:\mathrm{still}\:\mathrm{different}\:\mathrm{from}\:\mathrm{yours}.\:\mathrm{I}\:\mathrm{get} \\ $$$$\frac{\mathrm{sin}\:\mathrm{51}°}{\mathrm{2sin}\:\mathrm{1}°}\approx\mathrm{22}.\mathrm{2647} \\ $$
Commented by hardmath last updated on 10/May/24
thank you dear professor
$$\mathrm{thank}\:\mathrm{you}\:\mathrm{dear}\:\mathrm{professor} \\ $$

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