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sin-2-1-o-sin-2-5-o-sin-2-9-o-sin-2-89-o-a-1-b-b-Help-me-please-




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

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