Verifying identities

[frauleinedoctor]

Ive run myself into a snowbank mathematically I feel....How do I fix this? :

Verify the identity.

csc^4 x - cot^4 x = 2 csc^2 x - 1 <-- verify this one

So I...

(csc^2 x - cot^2 x ) (csc^2x + cot^2 x ) =

(1/sin^2 x - cos^2 x / sin^2 x ) ( 1/sin^2 x + cos^2 x / sin^2 x ) =

( (1 - cos^2 x)/ sin^2 x ) (( 1 + cos^2) / sin^2 x ) =

(sin^2 x / sin^2 x ) (( 1 + cos^2 x ) / sin^2 x ) =


Am I on the right track here?

 

[mmm4444bot]

… Am I on the right track here?

It looks good to me, but I only glanced over your work after approaching it differently, myself. I reached the stage at which you stopped.

Clearly, you must know that sin^2/sin^2 is 1, but you did not make that simplification. It appears that you stopped abrubtly. (Snowbanks tend to make that happen, if you don't keep in mind what's coming up.)

Were you concerned that your efforts might be wasted, if you were to have continued trying?

If so, that's ironic because you're only a couple steps away from finishing.

When I see these types of exercises with inverse functions, I like to make my first step rewriting everything in terms of the sine and cosine functions. (This is not necessary; I'm just showing you how I got to where you stopped.)

I grossly abbreviate the following. My true motivation for this is pure laziness, but I'm going to spin it to you as a magnanimous effort for you to see my steps "more clearly". (Pssst. Don't tell Denis.)

csc(x)^4 - cot(x)^4 = 2 * csc(x)^4 - 1

Rewrite CSC as 1/S and COT as C/S:

1/S^4 - C^4/S^4 = 2/S^2 - 1

(1 - C^4)/S^4 = (2 - S^2)/S^2

(1 + C^2)/S^2 * (1 - C^2)/S^2 = (2 - S^2)/S^2

(1 + C^2)/S^2 * S^2/S^2 = (2 - S^2)/S^2

(The line above is where you stopped; I continue by simplifying S^2/S^2.)

(1 + C^2)/S^2 = (2 - S^2)/S^2

We note that the denominator is now the same on both sides, so we only need to show that the numerators are equal.

1 + C^2 = 2 - S^2

Now, what do you supposed C^2 is equal to? Well, you can find out using the Pythagorean Identity, right?

Once you find an expression for C^2, then substitute it in, and see what happens.

 

[soroban]

Hello, frauleinedoctor!

You should become familiar with the "other" identities and their variations:

. . \(\displaystyle \begin{array}{ccccccc}\sec^2\!x \;=\; \tan^2\!x + 1 & \Rightarrow & \sec^2\!x -1 \;=\; \tan^2\!x & \Rightarrow & \sec^2\!x - \tan^2\!x \;=\; 1 \\ \csc^2x \;=\; \cot^2\!x + 1 & \Rightarrow & \csc^2\!x - 1 \;=\; \cot^2\!x & \Rightarrow & \csc^2\!x - \cot^2\!x \;=\; 1 \end{array}\)

Verify the identity: .\(\displaystyle \csc^4\!x - \cot^4\!x \:=\: 2\csc^2\!x - 1\)

\(\displaystyle \text{On the left, we have: }\;\underbrace{(\csc^2\!x - \cot^2\!x)}\,(\csc^2\!x + \underbrace{\cot^2\!x})\)
. . . . . . . . . . . . . . . . . .\(\displaystyle ^{\text{This is 1}}\). . . . . . \(\displaystyle ^{\text{This is }\csc^2\!x-1}\)


\(\displaystyle \text{Therefore: }\;1\cdot(\csc^2\!x + \csc^2\!x - 1) \;=\;2\csc^2\!x - 1\)