Understanding Trigonometric Identities: Solving for -1

[Marcus27]

Homework Statement

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

Homework Equations

Sin^2 x + cos^2 x == 1

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?. [/B]

 

[SteamKing]

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.

 

[Marcus27]

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.

Thanks, that puts my mind at ease.

 

[HallsofIvy]

I, for example, would, seeing that "[itex]sin^4(x)[/itex]" change everything else to "sin(x)".
Since [itex]cos^2(x)= 1- sin^2(x)[/itex], the numerator is [itex]sin^4(x)+ sin^2(x)(1- sin^2(x))= sin^4()+ sin^2(x)- sin^4(x)= sin^2(x)[/itex]. Is that what your textbook does?