Tính giá trị biểu thức $A={\sin }^2{10}^{°}+{\sin }^2{20}^{°}+\dots +{\sin }^2{180}^{°}$

Chọn A

Ta có \(\cot \alpha  =  - \frac{1}{2} \Rightarrow \tan \alpha  =  - 2\).

\(1 + {\tan ^2}\alpha  = \frac{1}{{{{\cos }^2}\alpha }} \Leftrightarrow {\cos ^2}\alpha  = \frac{1}{{1 + {{\tan }^2}\alpha }} = \frac{1}{{1 + {{\left( { - 2} \right)}^2}}} = \frac{1}{5}\).

Suy ra \(\cos \alpha  =  \pm \frac{{\sqrt 5 }}{5}\).

a)\({\left( {{{\sin }^2}\alpha  + {{\cos }^2}\alpha } \right)^2} = 1\)

b) \(1 + 2\sin \alpha  \cdot \cos \alpha  = {\sin ^2}\alpha  + {\cos ^2}\alpha  + 2\sin \alpha  \cdot \cos \alpha  = {(\sin \alpha  + \cos \alpha )^2}\).

c) \(1 - 2\sin \alpha  \cdot \cos \alpha  = {\sin ^2}\alpha  + {\cos ^2}\alpha  - 2\sin \alpha  \cdot \cos \alpha  = {(\sin \alpha  - \cos \alpha )^2}\).

d) Ta có: \({\sin ^4}\alpha  + {\cos ^4}\alpha  + 2{\sin ^2}\alpha  \cdot {\cos ^2}\alpha  = {\left( {{{\sin }^2}\alpha  + {{\cos }^2}\alpha } \right)^2} = 1\).

Do đó: \(1 - 2{\sin ^2}\alpha {\cos ^2}\alpha  = {\sin ^4}\alpha  + {\cos ^4}\alpha \).