Question Number 212186 by Ismoiljon_008 last updated on 05/Oct/24
$$ \\ $$$$\:\:\:{sin}^{\mathrm{2}} \:\mathrm{1}^{{o}} \:+\:{sin}^{\mathrm{2}} \:\mathrm{5}^{{o}} \:+\:{sin}^{\mathrm{2}} \:\mathrm{9}^{{o}} \:+\:...\:{sin}^{\mathrm{2}} \:\mathrm{89}^{{o}} \:=\:{a}\:\frac{\mathrm{1}}{{b}} \\ $$$$\:\:\:{b}\:=\:? \\ $$$$\:\:\:\mathbb{H}{elp}\:{me},\:{please} \\ $$$$ \\ $$
Answered by mehdee7396 last updated on 05/Oct/24
$${s}^{\mathrm{2}} \mathrm{1}+{s}^{\mathrm{2}} \mathrm{5}+{s}^{\mathrm{2}} \mathrm{9}+...+{s}^{\mathrm{2}} \mathrm{41}+{s}^{\mathrm{2}} \mathrm{45}+{c}^{\mathrm{2}} \mathrm{41}+...+{c}^{\mathrm{2}} \mathrm{9}+{c}^{\mathrm{2}} \mathrm{5}+{c}^{\mathrm{2}} \mathrm{1}= \\ $$$$\mathrm{11}+\frac{\mathrm{1}}{\mathrm{2}}={a}\frac{\mathrm{1}}{{b}}\Rightarrow{b}=\mathrm{2}\:\checkmark \\ $$
Commented by Ismoiljon_008 last updated on 05/Oct/24
$$\:\:\:\mathscr{T}\:{hank}\:{you}\:{very}\:{much} \\ $$