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Question Number 221939 by Ismoiljon_008 last updated on 13/Jun/25
sin^2 1° + sin^2 3° + sin^2 5° + ... + sin^2 269° = ? Help me, please
$$ \\ $$$$\:\:\:{sin}^{\mathrm{2}} \mathrm{1}°\:+\:{sin}^{\mathrm{2}} \mathrm{3}°\:+\:{sin}^{\mathrm{2}} \mathrm{5}°\:+\:…\:+\:{sin}^{\mathrm{2}} \mathrm{269}°\:=\:? \\ $$$$\:\:\:\mathcal{H}{elp}\:{me},\:\:{please} \\ $$$$ \\ $$
Commented by Rasheed.Sindhi last updated on 14/Jun/25
sin^2 θ+sin^2 (270−θ)=1
$$\mathrm{sin}^{\mathrm{2}} \theta+\mathrm{sin}^{\mathrm{2}} \left(\mathrm{270}−\theta\right)=\mathrm{1}\:\: \\ $$
Answered by wewji12 last updated on 13/Jun/25
Σ_(h=1) ^N sin^2 ((π/(180))h)=−Σ_(h=0) ^N ( ((e^(−(π/(180))ih) −e^((π/(180))iπh) )/2))^2 =Σ_(h=0) ^N (−(1/4)e^(−(1/(90))πh) −(1/4)e^((1/(90))πh) +(1/2))= −(1/4)Σ e^(−(1/(90))πh) −(1/4)Σ e^((1/(90))πh) +(1/2)N S(N)=−(1/4)∙(((^(90) (√e))/(^(90) (√e)−1))−((^(90) (√e^(−N) ))/(^(90) (√e)−1)))−(1/4)(((^(90) (√e^(N+1) )−1)/(^(90) (√e)−1)))+(1/2)N N=269 S(269)=134.5
$$\underset{{h}=\mathrm{1}} {\overset{{N}} {\sum}}\:\mathrm{sin}^{\mathrm{2}} \left(\frac{\pi}{\mathrm{180}}{h}\right)=−\underset{{h}=\mathrm{0}} {\overset{{N}} {\sum}}\:\left(\:\frac{{e}^{−\frac{\pi}{\mathrm{180}}\boldsymbol{{i}}{h}} −{e}^{\frac{\pi}{\mathrm{180}}\boldsymbol{{i}}\pi{h}} }{\mathrm{2}}\right)^{\mathrm{2}} \\ $$$$=\underset{{h}=\mathrm{0}} {\overset{{N}} {\sum}}\left(−\frac{\mathrm{1}}{\mathrm{4}}{e}^{−\frac{\mathrm{1}}{\mathrm{90}}\pi{h}} −\frac{\mathrm{1}}{\mathrm{4}}{e}^{\frac{\mathrm{1}}{\mathrm{90}}\pi{h}} +\frac{\mathrm{1}}{\mathrm{2}}\right)= \\ $$$$−\frac{\mathrm{1}}{\mathrm{4}}\Sigma\:{e}^{−\frac{\mathrm{1}}{\mathrm{90}}\pi{h}} −\frac{\mathrm{1}}{\mathrm{4}}\Sigma\:{e}^{\frac{\mathrm{1}}{\mathrm{90}}\pi{h}} +\frac{\mathrm{1}}{\mathrm{2}}{N} \\ $$$${S}\left({N}\right)=−\frac{\mathrm{1}}{\mathrm{4}}\centerdot\left(\frac{\:^{\mathrm{90}} \sqrt{{e}}}{\:^{\mathrm{90}} \sqrt{{e}}−\mathrm{1}}−\frac{\:^{\mathrm{90}} \sqrt{{e}^{−{N}} }}{\:^{\mathrm{90}} \sqrt{{e}}−\mathrm{1}}\right)−\frac{\mathrm{1}}{\mathrm{4}}\left(\frac{\:^{\mathrm{90}} \sqrt{{e}^{{N}+\mathrm{1}} }−\mathrm{1}}{\:^{\mathrm{90}} \sqrt{{e}}−\mathrm{1}}\right)+\frac{\mathrm{1}}{\mathrm{2}}{N} \\ $$$${N}=\mathrm{269} \\ $$$${S}\left(\mathrm{269}\right)=\mathrm{134}.\mathrm{5} \\ $$
Commented by Ismoiljon_008 last updated on 13/Jun/25
thank you
$${thank}\:{you} \\ $$$$ \\ $$
Answered by Rasheed.Sindhi last updated on 16/Jun/25
sin^2 1° + sin^2 3° + sin^2 5° + ... + sin^2 269° = ? •sin^2 θ+sin^2 (270−θ) = sin^2 θ+(sin 270 cos θ − cos 270 sin θ)^2 = sin^2 θ+((−1) cos θ − (0) sin θ)^2 =sin^2 θ+cos^2 θ =1 sin^2 1°+sin^2 269°=1 sin^2 3°+sin^2 267°=1 sin^2 5°+sin^2 265°=1 ... ... sin^2 135°+sin^2 135°=1 _ sin^2 1°+sin^2 3°+...+sin^2 269°−sin^2 135 =1×68−(0.5) =67.5
$$\:\:\:{sin}^{\mathrm{2}} \mathrm{1}°\:+\:{sin}^{\mathrm{2}} \mathrm{3}°\:+\:{sin}^{\mathrm{2}} \mathrm{5}°\:+\:…\:+\:{sin}^{\mathrm{2}} \mathrm{269}°\:=\:? \\ $$$$\bullet\mathrm{sin}^{\mathrm{2}} \theta+\mathrm{sin}^{\mathrm{2}} \left(\mathrm{270}−\theta\right) \\ $$$$=\:\mathrm{sin}^{\mathrm{2}} \theta+\left(\mathrm{sin}\:\mathrm{270}\:\mathrm{cos}\:\theta\:−\:\mathrm{cos}\:\mathrm{270}\:\:\mathrm{sin}\:\theta\right)^{\mathrm{2}} \\ $$$$=\:\mathrm{sin}^{\mathrm{2}} \theta+\left(\left(−\mathrm{1}\right)\:\mathrm{cos}\:\theta\:−\:\left(\mathrm{0}\right)\:\mathrm{sin}\:\theta\right)^{\mathrm{2}} \\ $$$$=\mathrm{sin}^{\mathrm{2}} \theta+\mathrm{cos}^{\mathrm{2}} \theta \\ $$$$=\mathrm{1} \\ $$$$\: \\ $$$$\:\mathrm{sin}^{\mathrm{2}} \:\mathrm{1}°+\mathrm{sin}^{\mathrm{2}} \:\mathrm{269}°=\mathrm{1} \\ $$$$\mathrm{sin}^{\mathrm{2}} \:\mathrm{3}°+\mathrm{sin}^{\mathrm{2}} \:\mathrm{267}°=\mathrm{1} \\ $$$$\mathrm{sin}^{\mathrm{2}} \:\mathrm{5}°+\mathrm{sin}^{\mathrm{2}} \:\mathrm{265}°=\mathrm{1} \\ $$$$… \\ $$$$… \\ $$$$\underline{\mathrm{sin}^{\mathrm{2}} \:\mathrm{135}°+\mathrm{sin}^{\mathrm{2}} \:\mathrm{135}°=\mathrm{1}\:\:\:\:\:\:\:_{\:} } \\ $$$$\mathrm{sin}^{\mathrm{2}} \:\mathrm{1}°+\mathrm{sin}^{\mathrm{2}} \:\mathrm{3}°+…+\mathrm{sin}^{\mathrm{2}} \:\mathrm{269}°−\mathrm{sin}^{\mathrm{2}} \mathrm{135}\:=\mathrm{1}×\mathrm{68}−\left(\mathrm{0}.\mathrm{5}\right) \\ $$$$=\mathrm{67}.\mathrm{5} \\ $$
Commented by mr W last updated on 16/Jun/25
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