Question

Tính  \( B = cos^21^0 +cos^22^0+...+cos^288^0+cos^289^0\) 

Answer 1

Ta có \(\cos1^o=\sin89^o\)

         \(\cos2^o=sin88^o\)

          ................

          \(\cos44^o=\sin46^o\)

           \(\cos45^o=\frac{\sqrt{2}}{2}\)

\(\Rightarrow\cos^21^o=\sin^289^o\)

     \(\cos^22^o=\sin^288^o\)

      ....................................

     \(\cos^244^o=\sin^246^o\)

     \(\cos^245^o=\frac{2}{4}=\frac{1}{2}\)

Khi đó \(B=\sin^289^o+\sin^288^o+...+\sin^246^o+\cos^245^o+\cos^246^o+...+\cos^289^o\)

\(=\left(\sin^289^o+\cos^289^o\right)+\left(\sin^288^o+\cos^288^o\right)+...+\left(\sin^246^o+\cos^246^o\right)+\cos^245^o\)

\(=1+1+...+1+\frac{1}{2}\)(44 số 1)

\(=44+\frac{1}{2}=\frac{89}{2}=44,5\)