How to prove that #(2sin(4theta)-sin(6theta)-sin(2theta))/(2sin(4theta)+sin(6theta)+sin(2theta)) = tan^2(theta)# ?

1 Answer
Oct 21, 2017

Use identities and axioms to make changes to only one side, until it is identical to the other side.

Explanation:

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