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Question Number 154640 by Study last updated on 20/Sep/21
Σ_1 ^(89) sin^2 (x)=?
$$\underset{\mathrm{1}} {\overset{\mathrm{89}} {\sum}}{sin}^{\mathrm{2}} \left({x}\right)=? \\ $$
Answered by qaz last updated on 20/Sep/21
Σ_(x=1°) ^(89°) sin^2 x=Σ_(x=1°) ^(89°) sin^2 (90°−x)=Σ_(x=1°) ^(89°) cos^2 x=(1/2)Σ_(n=1) ^(89) =((89)/2)
$$\underset{\mathrm{x}=\mathrm{1}°} {\overset{\mathrm{89}°} {\sum}}\mathrm{sin}\:^{\mathrm{2}} \mathrm{x}=\underset{\mathrm{x}=\mathrm{1}°} {\overset{\mathrm{89}°} {\sum}}\mathrm{sin}\:^{\mathrm{2}} \left(\mathrm{90}°−\mathrm{x}\right)=\underset{\mathrm{x}=\mathrm{1}°} {\overset{\mathrm{89}°} {\sum}}\mathrm{cos}\:^{\mathrm{2}} \mathrm{x}=\frac{\mathrm{1}}{\mathrm{2}}\underset{\mathrm{n}=\mathrm{1}} {\overset{\mathrm{89}} {\sum}}=\frac{\mathrm{89}}{\mathrm{2}} \\ $$
Commented by Study last updated on 20/Sep/21
why Σ_(x=1) ^(89) cos^2 x equal to (1/2)Σ_(n=1) ^(89) ?
$${why}\:\underset{{x}=\mathrm{1}} {\overset{\mathrm{89}} {\sum}}{cos}^{\mathrm{2}} {x}\:\:\:{equal}\:{to}\:\:\frac{\mathrm{1}}{\mathrm{2}}\underset{{n}=\mathrm{1}} {\overset{\mathrm{89}} {\sum}}\:? \\ $$
Commented by ARUNG_Brandon_MBU last updated on 20/Sep/21
S=Σ_(x=1°) ^(89) sin^2 x...eqn(i) =Σ_(x=1°) ^(89) sin^2 (90°−x) =Σ_(x=1) ^(89) cos^2 x...eqn(ii) eqn(i)+eqn(ii) give; 2S=Σ_(x=1°) ^(89) sin^2 x+Σ_(x=1°) ^(89) cos^2 x =Σ_(x=1°) ^(89) (sin^2 x+cos^2 x)=Σ_(x=1°) ^(89) (1) ⇒S=(1/2)Σ_(x=1°) ^(89) (1)
$${S}=\underset{{x}=\mathrm{1}°} {\overset{\mathrm{89}} {\sum}}\mathrm{sin}^{\mathrm{2}} {x}…\mathrm{eqn}\left({i}\right) \\ $$$$\:\:\:=\underset{{x}=\mathrm{1}°} {\overset{\mathrm{89}} {\sum}}\mathrm{sin}^{\mathrm{2}} \left(\mathrm{90}°−{x}\right) \\ $$$$\:\:\:=\underset{{x}=\mathrm{1}} {\overset{\mathrm{89}} {\sum}}\mathrm{cos}^{\mathrm{2}} {x}…\mathrm{eqn}\left({ii}\right) \\ $$$$\mathrm{eqn}\left({i}\right)+\mathrm{eqn}\left({ii}\right)\:\mathrm{give}; \\ $$$$\mathrm{2}{S}=\underset{{x}=\mathrm{1}°} {\overset{\mathrm{89}} {\sum}}\mathrm{sin}^{\mathrm{2}} {x}+\underset{{x}=\mathrm{1}°} {\overset{\mathrm{89}} {\sum}}\mathrm{cos}^{\mathrm{2}} {x} \\ $$$$\:\:\:\:\:\:=\underset{{x}=\mathrm{1}°} {\overset{\mathrm{89}} {\sum}}\left(\mathrm{sin}^{\mathrm{2}} {x}+\mathrm{cos}^{\mathrm{2}} {x}\right)=\underset{{x}=\mathrm{1}°} {\overset{\mathrm{89}} {\sum}}\left(\mathrm{1}\right) \\ $$$$\:\:\:\:\:\:\Rightarrow{S}=\frac{\mathrm{1}}{\mathrm{2}}\underset{{x}=\mathrm{1}°} {\overset{\mathrm{89}} {\sum}}\left(\mathrm{1}\right) \\ $$
Answered by mr W last updated on 20/Sep/21
sin^2 1°+sin^2 2°+...+sin^2 44°+sin^2 45°+sin^2 46°+...+sin^2 88°+sin^2 89° =(sin^2 1°+sin^2 89°)+(sin^2 2°+sin^2 88°)+...+(sin^2 44°+sin^2 46°)+sin^2 45° =(sin^2 1°+cos^2 1°)+(sin^2 2°+cos^2 2°)+...+(sin^2 44°+cos^2 44°)+sin^2 45° =1+1+...+1+((1/( (√2))))^2 =44+(1/2) =((89)/2)
$$\mathrm{sin}^{\mathrm{2}} \:\mathrm{1}°+\mathrm{sin}^{\mathrm{2}} \:\mathrm{2}°+…+\mathrm{sin}^{\mathrm{2}} \:\mathrm{44}°+\mathrm{sin}^{\mathrm{2}} \:\mathrm{45}°+\mathrm{sin}^{\mathrm{2}} \:\mathrm{46}°+…+\mathrm{sin}^{\mathrm{2}} \:\mathrm{88}°+\mathrm{sin}^{\mathrm{2}} \:\mathrm{89}° \\ $$$$=\left(\mathrm{sin}^{\mathrm{2}} \:\mathrm{1}°+\mathrm{sin}^{\mathrm{2}} \:\mathrm{89}°\right)+\left(\mathrm{sin}^{\mathrm{2}} \:\mathrm{2}°+\mathrm{sin}^{\mathrm{2}} \:\mathrm{88}°\right)+…+\left(\mathrm{sin}^{\mathrm{2}} \:\mathrm{44}°+\mathrm{sin}^{\mathrm{2}} \:\mathrm{46}°\right)+\mathrm{sin}^{\mathrm{2}} \:\mathrm{45}° \\ $$$$=\left(\mathrm{sin}^{\mathrm{2}} \:\mathrm{1}°+\mathrm{cos}^{\mathrm{2}} \:\mathrm{1}°\right)+\left(\mathrm{sin}^{\mathrm{2}} \:\mathrm{2}°+\mathrm{cos}^{\mathrm{2}} \:\mathrm{2}°\right)+…+\left(\mathrm{sin}^{\mathrm{2}} \:\mathrm{44}°+\mathrm{cos}^{\mathrm{2}} \:\mathrm{44}°\right)+\mathrm{sin}^{\mathrm{2}} \:\mathrm{45}° \\ $$$$=\mathrm{1}+\mathrm{1}+…+\mathrm{1}+\left(\frac{\mathrm{1}}{\:\sqrt{\mathrm{2}}}\right)^{\mathrm{2}} \\ $$$$=\mathrm{44}+\frac{\mathrm{1}}{\mathrm{2}} \\ $$$$=\frac{\mathrm{89}}{\mathrm{2}} \\ $$

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