2019-05-01T04:18:43-04:00
cos^3a sin3a+sin^3a cos3a=3/4sin4a
1
2019-05-02T02:36:57-0400
sin 3 α cos ( 3 α ) + cos 3 α sin ( 3 α ) = \text{sin}^3\alpha \text{cos}(3\alpha) + \text{cos}^3\alpha \text{sin}(3\alpha)= sin 3 α cos ( 3 α ) + cos 3 α sin ( 3 α ) =
= sin 3 α ( 4 cos 3 α − 3 cos α ) + cos 3 α ( 3 sin α − 4 sin 3 α ) = =\text{sin}^3\alpha(4\text{cos}^3\alpha-3\text{cos}\alpha)+\text{cos}^3\alpha(3\text{sin}\alpha-4\text{sin}^3\alpha)= = sin 3 α ( 4 cos 3 α − 3 cos α ) + cos 3 α ( 3 sin α − 4 sin 3 α ) =
= 4 sin 3 α cos 3 α − 3 sin 3 α cos α + 3 cos 3 α sin α − 4 cos 3 α sin 3 α = =4\text{sin}^3\alpha\text{cos}^3\alpha-3\text{sin}^3\alpha\text{cos}\alpha+3\text{cos}^3\alpha\text{sin}\alpha-4\text{cos}^3\alpha\text{sin}^3\alpha= = 4 sin 3 α cos 3 α − 3 sin 3 α cos α + 3 cos 3 α sin α − 4 cos 3 α sin 3 α =
= 3 sin α cos α ( cos 2 α − sin 2 α ) = =3\text{sin}\alpha\text{cos}\alpha(\text{cos}^2\alpha-\text{sin}^2\alpha)= = 3 sin α cos α ( cos 2 α − sin 2 α ) =
= 3 2 sin ( 2 α ) cos ( 2 α ) = 3 4 sin ( 4 α ) =\frac{3}{2}\text{sin}(2\alpha)\text{cos}(2\alpha)=\frac{3}{4}\text{sin}(4\alpha) = 2 3 sin ( 2 α ) cos ( 2 α ) = 4 3 sin ( 4 α )
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