What is #cos^3theta-6sin^3theta# in terms of non-exponential trigonometric functions?

1 Answer
George C.
Dec 31, 2015

#cos^3(theta)-6 sin^3(theta)#

#=1/4(cos(3theta)+3cos(theta))+3/2(sin(3theta)-3sin(theta))#

Explanation:

Since this is of degree #3#, the place to start is the expansion of #cos(3 theta)# and #sin(3 theta)#.

By De Moivre's formula:

#cos(3 theta) + i sin(3 theta)#

#=(cos(theta)+i sin(theta))^3#

#=cos^3(theta)+3 cos^2(theta) sin(theta) i - 3 cos(theta)sin^2(theta) - sin^3(theta) i#

#=(cos^3(theta)-3cos(theta)sin^2(theta))-(sin^3(theta)-3sin(theta)cos^2(theta)) i#

#=(4cos^3(theta)-3cos(theta)) + (3sin(theta)-4sin^3(theta))i#

So equating Real and Imaginary parts we find:

#cos(3theta) = 4cos^3(theta)-3cos(theta)#

#sin(3theta) = 3sin(theta)-4sin^3(theta)#

Hence:

#cos^3(theta) = 1/4(cos(3theta)+3cos(theta))#

#sin^3(theta) = 1/4(3sin(theta)-sin(3theta))#

So:

#cos^3(theta)-6 sin^3(theta)#

#=1/4(cos(3theta)+3cos(theta))+3/2(sin(3theta)-3sin(theta))#

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