Question

If tan x=-4 and x is negative, find sin2x, cos2x and tan2x

Answer 1

So many ways to do this...

Answer 2

well i'm looking for the answer that uses double angle identities

Answer 3

\[\sin(2x) = \dfrac{2\tan(x)}{1 + \tan^2(x)}\]\[\cos(2x) = \dfrac{1 - \tan^2(x)}{1 + \tan^2(x)}\]

Answer 4

And\[\tan(2x) = \dfrac{2\tan (x)}{1 - \tan^2(x)}\]By dividing those two.

Answer 5

are those considered identities? I only have identities for \[\cos 2x=\cos ^{2}x-\sin ^{2}x\]

Answer 6

and \[\sin2x=2sinxcosx\]

Answer 7

Yes, these are identities.

Answer 8

ok thanks. I don't think we've gotten that far yet. Appreciate the help!!

Answer 9

OK, I'll try to help you using the identities you already know.

Answer 10

You do know the identity for \(\tan(2x)\) though, right?

Answer 11

yes that one was familiar and in my textbook in this chapter.

Answer 12

OK, you can find tan(2x) to begin with then.

Answer 13

Thats my work. maybe it will help someone else. Thanks again!!

Answer 14

OK, that's a valid approach too. Good going.

Answer 15

Just a little warning of sorts, though: even though your approach works here, tan(x) = -4 can also mean that sin(x) = 4 and cos(x) = -1. So you should always think twice before proceeding with these things.

Answer 16

definitely. I was told that SinX was negative in my book and i wrote it down of sorts but it wasn't very clear in the question i typed on this site. I'll be sure to keep an eye out!

Answer 17

Oh, that was given? It's a very vital piece of information.

Answer 18

hehe yea, I wrote "X is negative" when i meant to write "SINX was negative.