If sinâ¡Î¸+cosâ¡Î¸=a and cosâ¡Î¸âsinâ¡Î¸=b, sinâ¡Î¸(sinâ¡Î¸âcosâ¡Î¸)+sin2â¡Î¸sin2â¡Î¸âcos2â¡Î¸+sin3â¡Î¸sin3â¡Î¸âcos3â¡Î¸+⦠is equal to
1âab1+ab
1âa23âa2
1âab1+ab+1âa23âa2
1+ab1âabâa2â13âa2
We have
sin2â¡Î¸+sin4â¡Î¸+sin6â¡Î¸+â¯.....âsinâ¡Î¸cosâ¡Î¸+sin2â¡Î¸cos2â¡Î¸+sin3â¡Î¸+cos3â¡Î¸+â¯
=sin2â¡Î¸1âsin2â¡Î¸âsinâ¡Î¸cosâ¡Î¸1âsinâ¡Î¸cosâ¡Î¸=1âcosâ¡2θ1+cosâ¡2θâsinâ¡2θ(2âsinâ¡2θ)=1âab1+ab+1âa23-a2