Question

Answer 1

a: \(A=\sin^220^0+\sin^2100^0+\sin^2140^0\)

\(=\sin^220^0+\sin^240^0+\sin^280^0\)

\(=\frac{1-cos\left(2\cdot20\right)}{2}+\frac{1-cos\left(2\cdot40\right)}{2}+\frac{1-cos\left(2\cdot80\right)}{2}\)

\(=\frac{1-cos40^0+1-cos80^0+1-cos160^0}{2}=\frac{3-\left(cos40^0+cos80^0+cos160^0\right)}{2}\)

\(=\frac{3-\left\lbrack2\cdot cos\left(\frac{80^0+40^0}{2}\right)\cdot cos\left(\frac{80^0-40^0}{2}\right)-cos20^0\right\rbrack}{2}=\frac{3-\left\lbrack2\cdot cos60^0\cdot cos20^0-cos20^0\right\rbrack}{2}\)

\(=\frac{3-\left\lbrack2\cdot\frac12\cdot cos20^0-cos20^0\right\rbrack}{2}=\frac32\)

b: \(B=cos^210^0+cos^2110^0+cos^2130^0\)

\(=cos^210^0+cos^2\left\lbrack90^0-\left(-20^0\right)\right\rbrack+cos^2\left\lbrack90^0-\left(-40^0\right)\right\rbrack\)

\(=cos^210^0+\sin^220^0+\sin^240^0\)

\(=\frac{1+cos20^0}{2}+\frac{1-cos40^0}{2}+\frac{1-cos80^0}{2}=\frac{3+cos20^0-cos40^0-cos80^0}{2}\)

\(=\frac{3+cos20^0-\left(cos80^0+cos40^0\right)}{2}=\frac{3+cos20^0-2\cdot cos\left(\frac{80^0+40^0}{2}\right)\cdot cos\left(\frac{80^0-40^0}{2}\right)}{2}\)

\(=\frac{3+cos20^0-2\cdot cos60^0\cdot cos20^0}{2}=\frac32\)