Help with trig identities

1. Feb 18, 2015

Marcus27

1. The problem statement, all variables and given/known data
Show that (sin^4 x + (sin^2 x * cos^2 x)) / (cos^2 x - 1) == -1
2. Relevant equations
Sin^2 x + cos^2 x == 1

3. The attempt at a solution
(sin ^4 x + (sin^2 x * cos^2 x)) / (cos^2 x - 1)
= ((sin^2 x)(sin^2 x) + (sin^2 x * cos^2 x)) / (cos^2 x - 1)
=((sin^2 x)(1 - cos^2 x) + (sin^2x * cos^2 x)) / (cos^2 x -1 )
= (sin^2 x - (sin^2 x * cos^2 x) + (sin^2 x * cos^2 x)) / (cos^2 x - 1 )
= (sin^2 x) / (cos^2 x - 1 )
= (1 - cos^2 x ) / (cos^2 x -1)
= -1

I think this is correct, but when I looked up the answer in the back of the textbook it showed completely different working using different substitutions. Did I make any mistakes? or are there two or more solutions to this problem?, if this is the case would I be marked down in an exam for using this method?.

2. Feb 18, 2015

SteamKing

Staff Emeritus
In working trig identity problems, there may be more than one valid substitution which can be used to obtain a simplification, especially with complicated or lengthy expressions.

If this were an exam exercise, no, you should not be penalized for using a valid method of simplification, even if it differs from a method preferred by the instructor.

3. Feb 18, 2015

Marcus27

Thanks, that puts my mind at ease.

4. Feb 22, 2015

HallsofIvy

Staff Emeritus
I, for example, would, seeing that "$sin^4(x)$" change everything else to "sin(x)".
Since $cos^2(x)= 1- sin^2(x)$, the numerator is $sin^4(x)+ sin^2(x)(1- sin^2(x))= sin^4()+ sin^2(x)- sin^4(x)= sin^2(x)$. Is that what your text book does?