Question

Tính   sin²15⁰ + sin²30⁰+ sin²40⁰ + sin²75⁰+ sin²60⁰+ sin²50⁰                              

Answer 1

\(=\left(sin^215^0+sin^275^0\right)+\left(sin^230^0+sin^260^0\right)+\left(sin^240^0+sin^250^0\right)\)

\(=\left(sin^215^0+cos^215^0\right)+\left(sin^230^0+cos^230^0\right)+\left(sin^240^0+cos^240^0\right)\)

=1+1+1

=3

Answer 2

\(sin^215^o+sin^230^o+sin^240^o+sin^275^o+sin^260^o+sin^250^o\\ =\left(sin^215^o+sin^275^o\right)+\left(sin^230^o+sin^260^o\right)+\left(sin^240^o+sin^250^o\right)\\ =1+1+1=3\)