Identity Help!

[gabby]

I need to prove that :

2(tanx-cotx)
tan²x-cot²x

is equal to:

sin2x




Using identities of course. I've done this problem over and over and can't prove it!

 

[gabby]

What I've gotten so far

2(tanx-cotx)
tan²x-cot²x ->


2(sinx-cosx)
cosx sinx
sin²x - cos²x
cos²x sin²x ->

2(sin²x-cos²x)
sinxcosx
sin⁴x - cos⁴x
sin²xcos²x ->


flipped the denominator and multiplied:


2(sin²x-cos²x)(sinxcosx)
(sin⁴x-cos⁴x) ->



sin2x(sin²x-cos²x)
(sin⁴x-cos⁴x)



Aaaand i don't see how that would work. Am I doing something wrong??

 

[srmichael]

So starting from where you stopped:

\(\displaystyle \displaystyle \frac{sin(2x)[sin^2(x)-cos^2(x)]}{sin^4(x)-cos^4(x)}\)

Then the denominator is just the difference of two squares so:

\(\displaystyle \displaystyle \frac{sin(2x)[sin^2(x)-cos^2(x)]}{[sin^2(x)-cos^2(x)][sin^2(x)+cos^2(x)]}\)

\(\displaystyle \displaystyle \frac{sin(2x)[sin^2(x)-cos^2(x)]}{[sin^2(x)-cos^2(x)](1)}\)

Then the \(\displaystyle \displaystyle sin^2(x)-cos^2(x)\) will cancel leaving you with \(\displaystyle \displaystyle sin(2x)\)

 

[soroban]

Hello, gabby!

\(\displaystyle \dfrac{2(\tan x - \cot x)}{\tan^2\!x - \cot^2\!x} \;=\;\dfrac{2\left(\dfrac{\sin x}{\cos x} - \dfrac{\cos x}{\sin x}\right)}{\dfrac{\sin^2x}{\cos^2x} - \dfrac{\cos^2\!x}{\sin^2\!x}} \;=\;\frac{\dfrac{2(\sin^2\!x-\cos^2\!x}{\sin x\cos x}}{\dfrac{\sin^4\!x - \cos^4\!x}{\sin^2\!x\cos^2\!x}} \)


\(\displaystyle \text{Flipped the denominator and multiplied: }\:\dfrac{2(\sin^2\!x - \cos^2\!x)}{\sin x\cos x} \cdot \dfrac{\sin^2\!x\cos^2\!x}{\sin^4\!x - \cos^4\!x}\)

. . \(\displaystyle =\;\dfrac{2\sin x\cos x(\sin^2\!x - \cos^2\!x)}{\cos^4\!x-\cos^4\!x} \;=\;\dfrac{\sin2x(\sin^2\!x - \cos^2\!x)}{\sin^4\!x - \cos^4\!x} \)


Aaaand i don't see how that would work. Am I doing something wrong? . No!

\(\displaystyle \text{Factor the denominator: }\:\dfrac{\sin2x\,(\sin^2\!x - \cos^2\!x)}{(\sin^2\!x - \cos^2\!x)\,(\sin^2\!x + \cos^2\!x)} \;=\) .\(\displaystyle \dfrac{\sin2x}{\underbrace{\sin^2\!x+\cos^2\!x}_{\text{This is 1}}} \;=\;\sin2x\)
Edit: too slow . . . again!

 

[Deleted member 4993]

I need to prove that :

2(tanx-cotx)
tan²x-cot²x

is equal to:

sin2x

Using identities of course. I've done this problem over and over and can't prove it!

Another way...

\(\displaystyle \dfrac{2[tan(x)-cot(x)]}{tan^2(x) - cot^2(x)}\)

\(\displaystyle = \ \dfrac{2[tan(x)-cot(x)]}{[tan(x) - cot(x)][tan(x) + cot(x)]}\)

\(\displaystyle = \ \dfrac{2}{tan(x) + cot(x)}\)

\(\displaystyle = \ \dfrac{2}{\frac{sin(x)}{cos(x)} + \frac{cos(x)}{sin(x)}}\)

\(\displaystyle = \ \dfrac{2\cdot sin(x)\cdot cos(x)}{sin^2(x) + cos^2(x)}\)

\(\displaystyle = \ sin(2x)\)

 

[srmichael]

Ahhh, so many roads that lead to the same destination.

I love tutotring this subject to my pre-calculus students. Identities is, without a doubt, the one subject when I hear this comment the most:

"When am I ever going to need to use this stuff in real life?"