Question Number 209706 by mr W last updated on 19/Jul/24
$${find}\:{the}\:{sum}\:{of}\: \\ $$$$\mathrm{sin}^{\mathrm{2}} \:\mathrm{1}°+\mathrm{sin}^{\mathrm{2}} \:\mathrm{2}°+…+\mathrm{sin}^{\mathrm{2}} \:\mathrm{60}°=? \\ $$
Answered by BaliramKumar last updated on 19/Jul/24
$$\underset{{x}=\mathrm{1}} {\overset{\mathrm{60}} {\sum}}\mathrm{sin}^{\mathrm{2}} {x}\:=\:\underset{{x}=\mathrm{1}} {\overset{\mathrm{29}} {\sum}}\mathrm{sin}^{\mathrm{2}} {x}\:+\mathrm{15}.\mathrm{5} \\ $$
Commented by mr W last updated on 19/Jul/24
$${and}\:\:\underset{{x}=\mathrm{1}} {\overset{\mathrm{29}} {\sum}}\mathrm{sin}^{\mathrm{2}} {x}=? \\ $$
Answered by MM42 last updated on 19/Jul/24
$${sin}^{\mathrm{2}} {a}=\frac{\mathrm{1}}{\mathrm{2}}\left(\mathrm{1}−{cos}\mathrm{2}{a}\right) \\ $$$$\Rightarrow{S}=\mathrm{30}−\frac{\mathrm{1}}{\mathrm{2}}\left({cos}\mathrm{2}+{cos}\mathrm{4}+…+{cos}\mathrm{120}\right) \\ $$$${s}={cos}\mathrm{2}+{cos}\mathrm{4}+…+{cos}\mathrm{120} \\ $$$$\Rightarrow\mathrm{2}{sin}\mathrm{1}×{s}=\mathrm{2}{sin}\mathrm{1}{cos}\mathrm{2}+\mathrm{2}{sin}\mathrm{1}{cos}\mathrm{4}+…+\mathrm{2}{sin}\mathrm{1}{cos}\mathrm{120} \\ $$$$=\left({sin}\mathrm{3}−{sin}\mathrm{1}\right)+\left({sin}\mathrm{5}−{sin}\mathrm{3}\right)+…+\left({sin}\mathrm{121}−{sin}\mathrm{119}\right) \\ $$$$={sin}\mathrm{121}−{sin}\mathrm{1}=\mathrm{2}{sin}\mathrm{60}{cos}\mathrm{61} \\ $$$$\Rightarrow{s}=\frac{\sqrt{\mathrm{3}}}{\mathrm{2}}×\frac{{cos}\mathrm{61}}{{sin}\mathrm{1}}\sim\:\mathrm{24}.\mathrm{06} \\ $$$$\Rightarrow{S}\sim\mathrm{30}−\mathrm{12}.\mathrm{03}=\mathrm{17}.\mathrm{7}\:\checkmark \\ $$$$ \\ $$
Commented by mr W last updated on 19/Jul/24
$${thanks}\:{sir}! \\ $$$${please}\:{recheck}. \\ $$$${S}=\mathrm{30}−\frac{\sqrt{\mathrm{3}}\:\mathrm{cot}\:\mathrm{1}°−\mathrm{3}}{\mathrm{8}}\approx\mathrm{17}.\mathrm{97} \\ $$
Commented by MM42 last updated on 19/Jul/24
$${cos}\mathrm{61}={cos}\mathrm{60}{cos}\mathrm{1}−{sin}\mathrm{1}{sin}\mathrm{60} \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}}\left({cos}\mathrm{1}−\sqrt{\mathrm{3}}{sin}\mathrm{1}\right) \\ $$$$\boldsymbol{\div}\mathrm{2}{sin}\mathrm{1}=\frac{\mathrm{1}}{\mathrm{4}}\left({cot}\mathrm{1}−\sqrt{\mathrm{3}}\right)\:\bigstar \\ $$