verify the identity

Question

(2cos^2 alpha-1)^2 /cos^4 alpha -sin^4 alpha = 1-2sin^2 alpha

Answer 1

Problem:

Verify the identity

(2cos^2 alpha-1)^2 /cos^4 alpha -sin^4 alpha = 1-2sin^2 alpha

Solution:

Let x = alpha

Rewrite as (note the added parentheses in the denominator needed to make this identity true. I graphed it.)

(2cos^2(x) - 1)^2 / (cos^4(x) - sin^4(x)) = 1 - 2*sin^2(x)

Double Angle Formulas:

a) cos(2x) = cos^2(x) - sin^2(x)

b) cos(2x) = 2*cos^2(x) - 1

c) cos(2x) = 1 - 2*sin^2(x)

Part 1:

Using formula b)

(2cos^2(x) - 1)^2

= [cos(2x)]^2

Part 2:

Using formula a) at the last line

cos^4(x) - sin^4(x)

= cos^2(x) * cos^2(x) - sin^2(x) * sin^2(x)

= cos^2(x) * (1 - sin^2(x)) - sin^2(x) * (1 - cos^2(x))

= cos^2(x) - sin^2(x)*cos^2(x) - sin^2(x) + sin^2(x) * cos^2(x)

= cos^2(x) - sin^2(x)

= cos(2x)

Part 3:

Using formula c)

1 - 2*sin^2(x)

= cos(2x)

Putting the first two together we get:

[cos(2x)]^2 / cos(2x)

[cos(2x) * cos^2(x)] / cos(2x) = cos(2x)

but this is the same as the third part.

cos(2x) = cos(2x)

Proven

Answer 2

Using the double angle identity cos(2α) = 2cos2(α) – 1 and factoring the difference of 2 squares in the denominator:

Using the double angle identity cos(2α) = cos2(α) – sin2(α) and the Pythagorean identity

cos2(α) + sin2(α) = 1:

Simplifying:

cos(2α) = 1 - 2sin²(α)

Then, finally using the other double angle cosine identity:

1 - 2sin²(α) = 1 - 2sin²(α)