You need the following identities:
1.cos2X+sin2X=12.sin(X+Y)=sinXcosY+sinYcosX3.cos(X−Y)=cosXcosY+sinXsinY4.sin(2X)=2∗sinXcosX
Solution:
cos3(A)cos(3A)+sin3(A)sin(3A)=cos2(A)(cosAcos(3A))+sin2(A)(sinAsin(3A))=
Formula 1 gives:
cos2(A)=1−sin2(A)
and
sin2(A)=1−cos2(A)
(1−sin2(A))(cos(A)cos(3A))+(1−cos2(A))(sin(A)sin(3A))=cos(A)cos(3A)+sin(A)sin(3A)−sin2(A)cos(A)cos(3A)−cos2(A)sin(A)sin(3A)=cos(A)cos(3A)+sin(A)sin(3A)−sin(A)cos(A)(sin(A)cos(3A)+sin(3A)cos(A))=
Formua 2 gives:
sin(A)cos(3A)+sin(3A)cos(A)=sin(4A) Formua 3 gives:
cos(A)cos(3A)+sin(A)sin(3A)=cos(2A)cos(2A)−sin(A)cos(A)(sin(4A))=
Formua 4 gives:
sin(4A)=2∗sin(2A)cos(2A)2∗sin(A)cos(A)=sin(2A)cos(2A)−2∗sin(A)cos(A)(sin(2A)cos(2A))=cos(2A)−sin(2A)sin(2A)cos(2A)=cos(2A)(1−sin2(2A))=
Using Formua 1 again :
cos(2A)(cos2(2A))=
cos3(2A)
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