3
$\begingroup$

My friend and I finished a test and we we're having a dispute over a question:

Ross was asked to find three different ways to express the identity below. How many examples are correct with appropriate restrictions? $$\frac{\cos⁡2x}{1-\sin^2⁡x}$$

Ex #1: $$\frac{\cot^2⁡x-1}{\cot^2⁡x}$$ Ex #2: $$\frac{2\cos^2⁡x-1}{\cos^2⁡x }$$ Ex #3: $$1-\tan^2⁡x$$

We both figured out that all three of the examples are the same. However, I said that three statements were correct while, he said two statements were correct. His justification for this was that for example one, $\cot^2x$ has the restrictions of $\frac{n\pi}{4}$ when it is in the denominator. I don't understand how that is possible and was wondering if someone could help me see his justification.

  • 1
    I have a trigonometry text right here that says an identity is an equation that is true for all replacements of variables in which both sides of the equation are defined. If this is your definition too, then the functions which are represented by these expressions need not have the same domain in order for it to be a identity. – 2017-01-13

1 Answers 1

2

To get to Ex #1 from the original expression, you divide the numerator and denominator by $\sin^{2}x$. Since the original denominator is $1 - \sin^{2}x = \cos^{2}x$, dividing by $\sin^{2}x$ removes integer multiples of $\pi$ from the natural domain. (For example, the original expression is defined at $x = 0$, but the expression in Ex #1 is not.)

By contrast:

  • Ex #2 differs from the original only by additive/subtractive trig identities (which hold everywhere).

  • Ex #3 differs from the original by canceling $\frac{\cos^{2}x}{\cos^{2}x} = 1$. This would change the domain if it weren't for the term $\tan^{2}x$ which is undefined precisely where $\cos^{2}x \neq 0$, which is precisely where the cancelation is correct.

  • 0
    Thank you Andrew this is exactly what I was looking for :) – 2017-01-13