Trigonometry question

[ImDaStyle]

---

Members don't see this ad.

This one problem is driving me insane, please help... =D

Verify!
cot^3x
-------- = cosx (csc^2x-1)
cscx

 

[bigstix808]

 

[Streetwolf]

This one problem is driving me insane, please help... =D

Verify!
cot^3x
-------- = cosx (csc^2x-1)
cscx

cot(x) = cos(x)/sin(x)

csc(x) = 1/sin(x)

So [cos^3(x) / sin^3(x)] / (1/sin(x)) = cos^3(x) / sin^2(x)

cos^3(x) = cos(x)cos^2(x) = cos(x)*[1-sin^2(x)] = cos(x) - cos(x)sin^2(x)

Divide all that by sin^2(x) and you get cos(x)/sin^2(x) - cos(x).

That is the same as cos(x) * [1/sin^2(x) - 1].

That is the same as cos(x) * [csc^2(x) - 1].

 

[ImDaStyle]

Yay ^__^ Thanks so much! I understand the problem at last xD!