Show that # sin^2x tan^2x = ( 3-4cos2x+cos4x ) / ( 4(1+cos2x) ) #?

1 Answer
Steve M
Oct 21, 2017

We will utilise the following identities:

# cos 2A -= 2cos^2A - 1 => cos^2A = 1/2(1+cos2A) #
# sin^2A + cos^2A -= 1 \ \ \ => sin^2A = 1/2(1-cos2A)#

Then:

# sin^2x tan^2x -= sin^2x sin^2x/cos^2x #

# " " = 1/2(1-cos2x) (1/2(1-cos2x)) / (1/2(1+cos2x)) #

# " " = 1/2(1-cos2x) (1-cos2x) / (1+cos2x) #

# " " = ( (1-cos2x) (1-cos2x) ) / ( 2(1+cos2x) ) #

# " " = ( 1-2cos2x+cos^2 2x ) / ( 2(1+cos2x) ) #

# " " = ( 1-2cos2x+1/2(1+cos4x) ) / ( 2(1+cos2x) ) #

# " " = ( 1/2(2-4cos2x+1+cos4x) ) / ( 2(1+cos2x) ) #
# " " = ( 3-4cos2x+cos4x ) / ( 4(1+cos2x) ) \ \ \ # QED

Related questions
See all questions in Proving Identities
Impact of this question
1338 views around the world
You can reuse this answer
Creative Commons License