How can I prove this equation is an identity? #sin^4w=1-cot^2w + cos^2w (cot^2w)/(csc^2w)#

2 Answers
Ananda Dasgupta
Mar 25, 2018

This is not an identity.

Explanation:

The left hand side is

#sin^4w = (1-cos^2w)^2=1-color(red)(2cos^2w)+cos^4w#

while the right hand side simplifies to

#1-cot^2w + cos^2w (cot^2w)/(csc^2w) = 1-cot^2w + cos^2w (cos^2w)/(sin^2wcsc^2w) = 1-color(red)(cot^2w)+cos^4w#

Since #cot^2w ne 2cos^2w# in general, this is not an identity.

Steve M
Mar 25, 2018

it is not an identity

Explanation:

We seek to prove:

# sin^4 w -= 1 - cot^2w + cos^2w (cot^2w/csc^2w )#

We can readily disprove the claim using a counter example:

Consider the case #w=pi/6#, then:

# LHS = sin^4 (pi/6) #
# \ \ \ \ \ \ \ \ = (1/2)^4 #
# \ \ \ \ \ \ \ \ = 1/16#

And:

# RHS = 1 - cot^2(pi/6) + cos^2(pi/6) cot^2(pi/6)/csc^2(pi/6)#

# \ \ \ \ \ \ \ \ = 1 - (sqrt(3))^2 + (sqrt(3)/2)^2 (sqrt(3))^2/(2)^2 #

# \ \ \ \ \ \ \ \ = 1 - 3 + (3/4) 3/4 #

# \ \ \ \ \ \ \ \ = -2 + 9/16 #

# \ \ \ \ \ \ \ \ = -23/16 #

# \ \ \ \ \ \ \ \ != LHS #

Hence this is not an identity

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