How do you simplify #f(theta)=sin4theta-cos6theta# to trigonometric functions of a unit #theta#?

1 Answer
Mar 15, 2018

#sin(theta)^6-15cos(theta)^2sin(theta)^4-4cos(theta)sin(theta)^3+15cos(theta)^4sin(theta)^2+4cos(theta)^3sin(theta)-cos(theta)^6#

Explanation:

We'll use the following two identities:
#sin(A+-B)=sinAcosB+-cosAsinB#
#cos(A+-B)=cosAcosB∓sinAsinB#

#sin(4theta)=2sin(2theta)cos(2theta)=2(2sin(theta)cos(theta))(cos^2(theta)-sin^2(theta))=4sin(theta)cos^3(theta)-4sin^3(theta)cos(theta)#

#cos(6theta)=cos^2(3theta)-sin^2(3theta)#
#=(cos(2theta)cos(theta)-sin(2theta)sin(theta))^2-(sin(2theta)cos(theta)+cos(2theta)sin(theta))^2#
#=(cos(theta)(cos^2(theta)-sin^2(theta))-2sin^2(theta)cos(theta))^2-(2cos^2(theta)sin(theta)+sin(theta)(cos^2(theta)-sin^2(theta))^2#
#=(cos^3(theta)-sin^2(theta)cos(theta)-2sin^2(theta)cos(theta))^2-(2cos^2(theta)sin(theta)+cos^2(theta)sin(theta)-sin^3(theta))^2#
#=(cos^3(theta)-3sin^2(theta)cos(theta))^2-(3cos^2(theta)sin(theta)-sin^3(theta))^2#
#=cos^6(theta)-6sin^2(theta)cos^4(theta)+9sin^4(theta)cos^2(theta)-9sin^2(theta)cos^4(theta)+6sin^4(theta)cos^2(theta)-sin^6(theta)#

#sin(4theta)-cos(6theta)=4sin(theta)cos^3(theta)-4sin^3(theta)cos(theta)-(cos^6(theta)-6sin^2(theta)cos^4(theta)+9sin^4(theta)cos^2(theta)-9sin^2(theta)cos^4(theta)+6sin^4(theta)cos^2(theta)-sin^6(theta))#
#=4sin(theta)cos^3(theta)-4sin^3(theta)cos(theta)-cos^6(theta)+6sin^2(theta)cos^4(theta)-9sin^4(theta)cos^2(theta)+9sin^2(theta)cos^4(theta)-6sin^4(theta)cos^2(theta)+sin^6(theta)#
#=sin(theta)^6-15cos(theta)^2sin(theta)^4-4cos(theta)sin(theta)^3+15cos(theta)^4sin(theta)^2+4cos(theta)^3sin(theta)-cos(theta)^6#