Question

How do you simplify `\frac { \cos ^ { 2} \alpha - \sin ^ { 2} \alpha } { \sin ^ { 4} \alpha - \cos ^ { 4} \alpha }`?

Answer 1

This equals `-1`

Explanation:

It's all about factoring the difference of squares.

`=((cosalpha + sin alpha)(cosalpha - sin alpha))/((sin^2alpha - cos^2alpha)(sin^2alpha + cos^2alpha))`

Because `sin^2x + cos^2x = 1`, we get:

`=((cosalpha + sin alpha)(cosalpha- sin alpha))/(sin^2alpha - cos^2alpha)`

`=((cosalpha + sin alpha)(cosalpha - sin alpha))/((sin alpha + cosalpha)(sin alpha - cosalpha)`

`= -1`

Hopefully this helps!

Answer 2

Given:

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

Factor the denominator, using the pattern (a^4-b^4) = (a^2-b^2)(a^2+b^2):

`(cos^2(alpha)-sin^2(alpha))/((cos^2(alpha)-sin^2(alpha))(cos^2(alpha)+sin^2(alpha))) =`

A factor in the denominator cancels the numerator:

`1/(cos^2(alpha)+sin^2(alpha)) =`

Use the identity `cos^2(x)+sin^2(x) = 1`

`1/1 = 1`