# Thread: Trig identities

1. ## Trig identities

14. a) Which equations are not identities? Justify your answers.
b) For those equations that are identities, state any restrictions on the variables.

i) $\displaystyle (1 - cos^2x)(1 - tan^2x) = \frac{sin^2x - 2sin^4x}{1 - sin^2x}$

Left Side
$\displaystyle = (1 - cos^2x)(1 - tan^2x)$
$\displaystyle = 1 - tan^2x - cos^2x + (-cos^2x)(-tan^2x)$
$\displaystyle = 1 - cos^2x - tan^2x + (cos^2x)(tan^2x)$
$\displaystyle = sin^2x - tan^2x + sin^2x$
$\displaystyle = (cos^2x)(tan^2x) - tan^2x + (cos^2x)(tan^2x)$
$\displaystyle = tan^2x(cos^2x - 1 + cos^2x)$
$\displaystyle = tan^2x(-sin^2x + cos^2x)$ Edit: Was an error made here?
$\displaystyle = tan^2x(-1)$
$\displaystyle = -tan^2x$

Right Side:
$\displaystyle = \frac{sin^2x - 2sin^4x}{1 - sin^2x}$
$\displaystyle = \frac{sin^2x - 2(sin^2x)(sin^2x)}{cos^2x}$
$\displaystyle = \frac{sin^2x(1 -2(1)(1))}{cos^2x}$
$\displaystyle = \frac{sin^2x(1 -2)}{cos^2x}$
$\displaystyle = \frac{sin^2x(-1)}{cos^2x}$
$\displaystyle = -tan^2x$

Left Side = Right Side
Therefore, this equation is an identity.

When I showed this solution to my teacher, I was told that had I used identities that we haven't learned yet. She said that on the exam, we were to only be able to use reciprocal identities and the Pythagorean identity $\displaystyle sin^2x + cos^2x = 1.$ She also told me that using these identities, I can make the $\displaystyle LS$ and $\displaystyle RS$ equal however with different values. Is there another way of proving this equation to be an identity?

Also, how do I know whether or not an equation will eventually end up as $\displaystyle LS = RS$? The only reason I know it with the few assigned questions is because there is an answer at the back of the book. Though on an exam I won't have access to these answers. Is it just trial and error? Because if it is, I can be going on for a pretty long time just trying to solve one question.

2. $\displaystyle (1-\cos^2{x})(1-\tan^2{x}) =$

$\displaystyle \sin^2{x}\left(1 - \frac{\sin^2{x}}{\cos^2{x}}\right) =$

$\displaystyle \sin^2{x} - \frac{\sin^4{x}}{\cos^2{x}} =$

$\displaystyle \frac{\sin^2{x}\cos^2{x}}{\cos^2{x}} - \frac{\sin^4{x}}{\cos^2{x}} =$

$\displaystyle \frac{\sin^2{x}\cos^2{x} - \sin^4{x}}{\cos^2{x}} =$

$\displaystyle \frac{\sin^2{x}(1 - \sin^2{x}) - \sin^4{x}}{1 - \sin^2{x}} =$

$\displaystyle \frac{\sin^2{x} - \sin^4{x} - \sin^4{x}}{1 - \sin^2{x}} =$

$\displaystyle \frac{\sin^2{x} - 2\sin^4{x}}{1 - \sin^2{x}}$

3. Originally Posted by RogueDemon
14. a) Which equations are not identities? Justify your answers.
b) For those equations that are identities, state any restrictions on the variables.

i) $\displaystyle (1 - cos^2x)(1 - tan^2x) = \frac{sin^2x - 2sin^4x}{1 - sin^2x}$

Left Side
$\displaystyle = (1 - cos^2x)(1 - tan^2x)$
$\displaystyle = 1 - tan^2x - cos^2x + (-cos^2x)(-tan^2x)$
$\displaystyle = 1 - cos^2x - tan^2x + (cos^2x)(tan^2x)$
$\displaystyle = sin^2x - tan^2x + sin^2x$
$\displaystyle = (cos^2x)(tan^2x) - tan^2x + (cos^2x)(tan^2x)$
$\displaystyle = tan^2x(cos^2x - 1 + cos^2x)$
$\displaystyle = tan^2x(-sin^2x + cos^2x)$ Edit: Was an error made here?
$\displaystyle = tan^2x(-1)$
$\displaystyle = -tan^2x$

Right Side:
$\displaystyle = \frac{sin^2x - 2sin^4x}{1 - sin^2x}$
$\displaystyle = \frac{sin^2x - 2(sin^2x)(sin^2x)}{cos^2x}$
$\displaystyle = \frac{sin^2x(1 -2(1)(1))}{cos^2x}$
$\displaystyle = \frac{sin^2x(1 -2)}{cos^2x}$
$\displaystyle = \frac{sin^2x(-1)}{cos^2x}$
$\displaystyle = -tan^2x$

Left Side = Right Side
Therefore, this equation is an identity.

When I showed this solution to my teacher, I was told that had I used identities that we haven't learned yet. She said that on the exam, we were to only be able to use reciprocal identities and the Pythagorean identity $\displaystyle sin^2x + cos^2x = 1.$ She also told me that using these identities, I can make the $\displaystyle LS$ and $\displaystyle RS$ equal however with different values. Is there another way of proving this equation to be an identity?

Also, how do I know whether or not an equation will eventually end up as $\displaystyle LS = RS$? The only reason I know it with the few assigned questions is because there is an answer at the back of the book. Though on an exam I won't have access to these answers. Is it just trial and error? Because if it is, I can be going on for a pretty long time just trying to solve one question.
When proving identities it is good practice to only change one side and manipulate it to get the other.

$\displaystyle (1 - cos^2x)(1 - tan^2x) = \frac{sin^2x - 2sin^4x}{1 - sin^2x}$

I have put my steps as spoilers along with explanation.

From the LHS

Use FOIL

Spoiler:
$\displaystyle 1-\tan^2(x) - \cos^2(x) + \sin^2(x)$

Multiply through by $\displaystyle \frac{\cos^2(x)}{\cos^2(x)}$ as this is the LCD of the expression.

Spoiler:
$\displaystyle \frac{\cos^2(x) - \sin^2(x) - \cos^4(x) + \sin^2(x) \cos^2(x)}{\cos^2(x)}$

Using the identity $\displaystyle \sin^2(x) + \cos^2(x) = 1$ rewrite any cos terms as sin.

Spoiler:
$\displaystyle \frac{(1-\sin^2(x)) - \sin^2(x) - (1-\sin^2(x))(1-\sin^2(x)) + \sin^2(x)(1-\sin^2(x))}{1- \sin^2(x)}$

Expand the brackets

Spoiler:
$\displaystyle \frac{1-2\sin^2(x) - 1 + 2\sin^2(x) - \sin^4(x) + \sin^2(x) - \sin^4(x)}{1-\sin^2(x)}$

Collect like terms and simplify

Spoiler:
$\displaystyle \frac{\sin^2(x) - 2\sin^4(x)}{1-\sin^2(x)}$

4. Hello, RogueDemon!

We should work on one side only
. . and try to make it equal the other side.

14. a) Which equations are not identities? Justify your answers.
. . .b) For those equations that are identities, state any restrictions on the variables.

$\displaystyle i)\;(1 - \cos^2\!x)(1 - \tan^2\!x) \:=\: \frac{\sin^2\!x - 2\sin^4\!x}{1 - \sin^2\!x}$

I started with the left side:

. . $\displaystyle (1-\cos^2\!x)(1-\tan^2\!x)$. . . . . . . . . .Replace $\displaystyle 1-\cos^2\!x$ with $\displaystyle \sin^2\!x$

. . . . . $\displaystyle =\;\sin^2\!x(1 - \tan^2\!x)$. . . . . . . . Replace $\displaystyle \tan^2\!x$ with $\displaystyle \frac{\sin^2\!x}{\cos^2\!x}$

. . . . . $\displaystyle =\;\sin^2\!x\left(1-\frac{\sin^2\!x}{\cos^2\!x}\right)$. . . . . . . Subtract the fractions

. . . . . $\displaystyle =\;\sin^2\!x\left(\frac{\cos^2\!x - \sin^2\!x}{\cos^2\!x}\right)$. . . . . Replace $\displaystyle \cos^2\!x$ with $\displaystyle 1-\sin^2\!x$

. . . . . $\displaystyle =\;\sin^2\!x\left(\frac{[1-\sin^2\!x]-\sin^2\!x}{\cos^2\!x}\right)$. . Combine like terms

. . . . . $\displaystyle =\;\sin^2\!x\left(\frac{1 - 2\sin^2\!x}{\cos^2\!x}\right)$. . . . . . .Multiply

. . . . . $\displaystyle =\;\frac{\sin^2\!x - 2\sin^4\!x}{\cos^2\!x}$. . . . . . . . . .Replace $\displaystyle \cos^2\!x$ with $\displaystyle 1-\sin^2\!x$

. . . . . $\displaystyle =\;\frac{\sin^2\!x-2\sin^4\!x}{1-\sin^2\!x}$

5. There is an error in the OP. For the LHS, you cannot go from

$\displaystyle =\tan^{2}(x)(-\sin^{2}(x)+\cos^{2}(x))$
$\displaystyle =\tan^{2}(x)(-1)$,

because it is only when the $\displaystyle \sin^{2}(x)$ and the $\displaystyle \cos^{2}(x)$ have the same sign, that they sum to 1.

Regards.