Prove:
#(2sin(4theta)-sin(6theta)-sin(2theta))/(2sin(4theta)+sin(6theta)+sin(2theta)) = tan^2(theta)#
Write #4theta# as #2theta + 2theta# and #6theta# as #2theta + 4theta#:
#(2sin(2theta+ 2theta)-sin(2theta+ 4theta)-sin(2theta))/(2sin(2theta+ 2theta)+sin(2theta+ 4theta)+sin(2theta)) = tan^2(theta)#
Substitute #sin(2theta+ 2theta) = 2sin(2theta)cos(2theta)#:
#(4sin(2theta)cos(2theta)-sin(2theta+ 4theta)-sin(2theta))/(4sin(2theta)cos(2theta)+sin(2theta+ 4theta)+sin(2theta)) = tan^2(theta)#
Substitute #sin(2theta+ 4theta) = sin(2theta)cos(4theta) + cos(2theta)sin(4theta)#:
#(4sin(2theta)cos(2theta)-(sin(2theta)cos(4theta) + cos(2theta)sin(4theta))-sin(2theta))/(4sin(2theta)cos(2theta)+(sin(2theta)cos(4theta) + cos(2theta)sin(4theta))+sin(2theta)) = tan^2(theta)#
Substitute #sin(4theta) = 2sin(2theta)cos(2theta)#:
#(4sin(2theta)cos(2theta)-(sin(2theta)cos(4theta) + 2cos^2(2theta)sin(2theta))-sin(2theta))/(4sin(2theta)cos(2theta)+(sin(2theta)cos(4theta) + 2cos^2(2theta)sin(theta))+sin(2theta)) = tan^2(theta)#
There is a common factor of #sin(2theta)# in both the numerator and denominator:
#(sin(2theta)(4cos(2theta)-(cos(4theta) + 2cos^2(2theta))-1))/(sin(2theta)(4cos(2theta)+(cos(4theta) + 2cos^2(2theta))+1)) = tan^2(theta)#
The common factor becomes 1:
#(4cos(2theta)-(cos(4theta) + 2cos^2(2theta))-1)/(4cos(2theta)+(cos(4theta) + 2cos^2(2theta))+1) = tan^2(theta)#
Distribute the -1 in the numerator and drop the () around the middler term in the denominator:
#(4cos(2theta)-cos(4theta) - 2cos^2(2theta)-1)/(4cos(2theta)+cos(4theta) + 2cos^2(2theta)+1) = tan^2(theta)#
Use the identity #cos(4theta) = 2cos^2(2theta) - 1#:
#(4cos(2theta)-2cos^2(2theta) + 1 - 2cos^2(2theta)-1)/(4cos(2theta)+2cos^2(2theta) - 1 + 2cos^2(2theta)+1) = tan^2(theta)#
Combine like terms:
#(4cos(2theta)-4cos^2(2theta))/(4cos(2theta)+4cos^2(2theta)) = tan^2(theta)#
A common factor of #4cos(2theta)# in both numerator and denominator cancels:
#(1-cos(2theta))/(1+cos(2theta)) = tan^2(theta)#
Use the identity #cos(2theta) = 1-2sin^2(theta)# in the numerator and #cos(2theta) = cos^2(theta)-1# in the denominator:
#(1-1 + 2sin^2(theta))/(1+2cos^2(theta)-1) = tan^2(theta)#
Combine like terms:
#(2sin^2(theta))/(2cos^2(theta)) = tan^2(theta)#
The 2s cancel and the left side becomes #tan^2(theta)#:
#tan^2(theta) = tan^2(theta)# Q.E.D