Question

simplify the expression: sin^2x-cos^2x/sin^2x-sinxcosx

Answer 1

Well lets work this out together. What would be the first step that you would take for wanting to solve something like this?

Answer 2

i know you have to use one of the fundamental indentities but i dont know where to start.

Answer 3

So when you have a problem that's a faction state. What 2 things stand out the most.

Answer 4

The \[ \sin^2- \cos^2 \] seems like I can use the Pythagorean identity, but the minus sign throws me a bit off. Also with the bottom \[\sin^2\] seems changable too.

Answer 5

Yea, it can throw you off but you have to think math wise. So if its not a + than work with the - the same way. So \[\sin ^2 -\cos ^2 = -1\].. And yea the bottom can be changed too.

Answer 6

Alrighty then. I don't really want to change the top to -1 since it seems a bit confusing with the denominator being all complex.

Answer 7

Here, something worked out. See if you understand this a bit.
sin2x sinx - cosx=0
--> [2sin(x)cos(x)]sin(x) - cos(x) = 0
--> 2sin^2(x)cos(x) - cos(x) = 0
--> 2[1 - cos^2(x)]cos(x) - cos(x) = 0
--> 2cos(x) - 2cos^3(x) - cos(x) = 0
--> 2cos^3(x) + cos(x) = 0

Answer 8

Wait. Okay. Which part are you simplifying? Sorry. I'm a bit slow.

Answer 9

This is the denominator.

Answer 10

Ohh. I get what you're trying to do, but I looked at the back of the book and it said the answer was 1 + cotx, and it looks so simple to get there, but my brain does not like to work with me.

Answer 11

I see why they would say that. Alright. Yea, cause cos x would be left cause it = 0. And the top I read my calc wrong I missed typed it was 1. So yes that is correct in the back of the book.

Answer 12

Ah. Okay. Well, I'm not too 100% on this, but I know a little bit more now. Thanks.

Answer 13

Yeah, no problem. If you need more help just send a holler.