Question #88936
cos^3a.cos 3a+sin^3a.sin3a=cos^3 2a
1
Expert's answer
2019-05-07T04:27:17-0400

You need the following identities: 


1.cos2X+sin2X=11. cos^2 X + sin^2 X = 12.sin(X+Y)=sinXcosY+sinYcosX2. sin(X+Y) = sin X cosY + sin Y cos X3.cos(XY)=cosXcosY+sinXsinY3. cos(X-Y) = cos X cos Y + sin X sin Y4.sin(2X)=2sinXcosX4. sin(2X)= 2*sin X cos X


Solution:


cos3(A)cos(3A)+sin3(A)sin(3A)=cos^3 (A) cos (3A) + sin^3 (A) sin (3A) =cos2(A)(cosAcos(3A))+sin2(A)(sinAsin(3A))=cos^2 (A) (cos A cos (3A)) + sin^2 (A)(sin A sin (3A)) =


Formula 1 gives:


cos2(A)=1sin2(A)cos^2 (A) = 1 - sin^2 (A)


and


sin2(A)=1cos2(A)sin^2 (A) = 1 - cos^2 (A)


(1sin2(A))(cos(A)cos(3A))+(1cos2(A))(sin(A)sin(3A))=(1-sin^2 (A)) (cos (A) cos (3A)) + (1-cos^2 (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^2 (A) cos (A) cos (3A) - cos^2 (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))=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)sin (A) cos (3A) + sin (3A) cos (A) = sin (4A)

Formua 3 gives:


cos(A)cos(3A)+sin(A)sin(3A)=cos(2A)cos (A) cos (3A) + sin (A) sin (3A) = cos (2A)cos(2A)sin(A)cos(A)(sin(4A))=cos (2A) - sin (A) cos (A) (sin (4A)) =


Formua 4 gives:


sin(4A)=2sin(2A)cos(2A)sin (4A) = 2*sin (2A) cos (2A)2sin(A)cos(A)=sin(2A)2*sin (A) cos (A) = sin (2A)cos(2A)2sin(A)cos(A)(sin(2A)cos(2A))=cos (2A) - 2*sin (A) cos (A) (sin (2A) cos (2A)) =cos(2A)sin(2A)sin(2A)cos(2A)=cos (2A) - sin (2A) sin (2A) cos( 2A) =cos(2A)(1sin2(2A))=cos (2A) (1 - sin^2 (2A)) =


Using Formua 1 again :


cos(2A)(cos2(2A))=cos( 2A) (cos^2 (2A)) =


cos3(2A)cos^3 (2A)

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