Bill D.

asked • 12/19/16# Prove the identity

(1 - cos2)(1 - tan2)

(1 - cos2)(1 - sin2/cos2)

1 - cos2 - sin2/cos2 + sin2

## 1 Expert Answer

On the right side denominator you probably means (1-sin2x)

but then it's not a trig identity. You can disprove it by showing one counterexample

Let x= 22.5 degrees, then the left side = 0 as (1-tan2x) = 1-tan45 = 1-1 = 0

but the right side does not equal zero

if you didn't leave out a parentheses on the right side, then

the right side becomes

sin2xcos2x-sin2x

let 2x = A, tan2x = sinA/cosA and expand the 3 factors on the left side

1-cosA - sinA/cosA + sinA = - sinA + sinAcosA

1-cosA - sinA/cosA +2sinA = sinAcosA

multiply by cosA

cosA -cos^2A -sinA +2sinAcosA = sinAcosA

cosA(1-cosA) -sinA(1-cosA) =0

factor out 1-cosA

(1-cosA)[cosA-sinA) = 0

set each factor = 0

1-cosA = 0

cosA = 1

A = 0

A=2x, x=A/2 = 0/2 = 0 +2npi or + n360 degrees where n= any integer

cosA-sinA = 0

cosA = sinA

A=2x

x = A/2 = 22 1/2 degrees or 112.5 degrees or pi/8 or 5pi/8

but when you check 22.5 degrees in the original given "identity" it doesn't work

in any event, it's not a trig identity. trig identities are true for all values of the variable. here only x=0 works

a trig identity is like cos^2 x+ sin^2x = 1, where it's true for all values of x

but this just shows it's not a trig identity, as it's not true for all x

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Gene G.

^{2}x)(1 - tan^{2}x) = sin^{2}x cos^{2}x/1-sin^{2}x12/19/16