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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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