Question Number 209706 by mr W last updated on 19/Jul/24
![find the sum of sin^2 1°+sin^2 2°+...+sin^2 60°=?](https://www.tinkutara.com/question/Q209706.png)
$${find}\:{the}\:{sum}\:{of}\: \\ $$$$\mathrm{sin}^{\mathrm{2}} \:\mathrm{1}°+\mathrm{sin}^{\mathrm{2}} \:\mathrm{2}°+…+\mathrm{sin}^{\mathrm{2}} \:\mathrm{60}°=? \\ $$
Answered by BaliramKumar last updated on 19/Jul/24
![Σ_(x=1) ^(60) sin^2 x = Σ_(x=1) ^(29) sin^2 x +15.5](https://www.tinkutara.com/question/Q209708.png)
$$\underset{{x}=\mathrm{1}} {\overset{\mathrm{60}} {\sum}}\mathrm{sin}^{\mathrm{2}} {x}\:=\:\underset{{x}=\mathrm{1}} {\overset{\mathrm{29}} {\sum}}\mathrm{sin}^{\mathrm{2}} {x}\:+\mathrm{15}.\mathrm{5} \\ $$
Commented by mr W last updated on 19/Jul/24
![and Σ_(x=1) ^(29) sin^2 x=?](https://www.tinkutara.com/question/Q209714.png)
$${and}\:\:\underset{{x}=\mathrm{1}} {\overset{\mathrm{29}} {\sum}}\mathrm{sin}^{\mathrm{2}} {x}=? \\ $$
Answered by MM42 last updated on 19/Jul/24
![sin^2 a=(1/2)(1−cos2a) ⇒S=30−(1/2)(cos2+cos4+...+cos120) s=cos2+cos4+...+cos120 ⇒2sin1×s=2sin1cos2+2sin1cos4+...+2sin1cos120 =(sin3−sin1)+(sin5−sin3)+...+(sin121−sin119) =sin121−sin1=2sin60cos61 ⇒s=((√3)/2)×((cos61)/(sin1))∼ 24.06 ⇒S∼30−12.03=17.7 ✓](https://www.tinkutara.com/question/Q209716.png)
$${sin}^{\mathrm{2}} {a}=\frac{\mathrm{1}}{\mathrm{2}}\left(\mathrm{1}−{cos}\mathrm{2}{a}\right) \\ $$$$\Rightarrow{S}=\mathrm{30}−\frac{\mathrm{1}}{\mathrm{2}}\left({cos}\mathrm{2}+{cos}\mathrm{4}+…+{cos}\mathrm{120}\right) \\ $$$${s}={cos}\mathrm{2}+{cos}\mathrm{4}+…+{cos}\mathrm{120} \\ $$$$\Rightarrow\mathrm{2}{sin}\mathrm{1}×{s}=\mathrm{2}{sin}\mathrm{1}{cos}\mathrm{2}+\mathrm{2}{sin}\mathrm{1}{cos}\mathrm{4}+…+\mathrm{2}{sin}\mathrm{1}{cos}\mathrm{120} \\ $$$$=\left({sin}\mathrm{3}−{sin}\mathrm{1}\right)+\left({sin}\mathrm{5}−{sin}\mathrm{3}\right)+…+\left({sin}\mathrm{121}−{sin}\mathrm{119}\right) \\ $$$$={sin}\mathrm{121}−{sin}\mathrm{1}=\mathrm{2}{sin}\mathrm{60}{cos}\mathrm{61} \\ $$$$\Rightarrow{s}=\frac{\sqrt{\mathrm{3}}}{\mathrm{2}}×\frac{{cos}\mathrm{61}}{{sin}\mathrm{1}}\sim\:\mathrm{24}.\mathrm{06} \\ $$$$\Rightarrow{S}\sim\mathrm{30}−\mathrm{12}.\mathrm{03}=\mathrm{17}.\mathrm{7}\:\checkmark \\ $$$$ \\ $$
Commented by mr W last updated on 19/Jul/24
![thanks sir! please recheck. S=30−(((√3) cot 1°−3)/8)≈17.97](https://www.tinkutara.com/question/Q209722.png)
$${thanks}\:{sir}! \\ $$$${please}\:{recheck}. \\ $$$${S}=\mathrm{30}−\frac{\sqrt{\mathrm{3}}\:\mathrm{cot}\:\mathrm{1}°−\mathrm{3}}{\mathrm{8}}\approx\mathrm{17}.\mathrm{97} \\ $$
Commented by MM42 last updated on 19/Jul/24
![cos61=cos60cos1−sin1sin60 =(1/2)(cos1−(√3)sin1) ÷2sin1=(1/4)(cot1−(√3)) ★](https://www.tinkutara.com/question/Q209729.png)
$${cos}\mathrm{61}={cos}\mathrm{60}{cos}\mathrm{1}−{sin}\mathrm{1}{sin}\mathrm{60} \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}}\left({cos}\mathrm{1}−\sqrt{\mathrm{3}}{sin}\mathrm{1}\right) \\ $$$$\boldsymbol{\div}\mathrm{2}{sin}\mathrm{1}=\frac{\mathrm{1}}{\mathrm{4}}\left({cot}\mathrm{1}−\sqrt{\mathrm{3}}\right)\:\bigstar \\ $$