Question

Prove the identity
(sinx)(tanx cosx - cotx cosx) = 1- 2 cos^2x

Answer 1

Rewrite \(\large \tan x\) and \(\large \cot x\) in terms of sines and cosines.
Then cancel some stuff out.
And simplify.\[\large \sin x\left(\tan x \cos x- \cot x \cos x\right) \qquad = \qquad \sin x\left(\frac{\sin x}{\cos x} \cos x- \frac{\cos x}{\sin x} \cos x\right)\]

Answer 2

how come this stuff never shows up when people use the symbols? I cant read it :(

Answer 3

i really appreciate your help, i just cant see anything

Answer 4

Hmm what browser are you using? :(

I think you'll have problems if you're on internet explorer.

Answer 5

oh that must be it! i am on that, i will switch quick and see if it shows up!
thank you so much!

Answer 6

WHUT! Why would you EFER use Internet Explorer? +_+
Is it your first day on the interwebs? :O

Answer 7

XD

Answer 8

bahahaha no, its just what ive always used. don't worry i get made fun of for using it all the time. and the computers at my school have that

Answer 9

fair enough :3

Answer 10

ok @zepdrix help! i went from (sinx)(tanx cosx - cotx cosx)
to (sinx)(sinx/cosx * cosx - cox/sinx *cosx)
to (sinx)(sinx - cos^2x/sinx)

NOW WHAT?

Answer 11

Hmm in the second set of brackets.. let's look at the first fraction there.
It looks like we have a couple of cosines that will cancel out, don't we?

Answer 12

Are my pretty graphics showing up now, or not so much? :o

Answer 13

kind of, but it is like cut off :/

Answer 14

Hehe yah, my bad.

Answer 15

\[\large \sin x\left(\frac{\sin x}{\cancel{\cos x}} \cancel{\cos x}- \frac{\cos x}{\sin x} \cos x\right)\]So we have a nice cancellation here right?

Answer 16

yep did that! then i multiplied cosx/sinx times cosx/1

Answer 17

\[\large \sin x\left(\sin x- \frac{\cos^2 x}{\sin x}\right)\]Hmm, Ok cool.
Next thing we'll want to do is `distribute` the \(\large \sin x\) to each term in the brackets.

Answer 18

which is sin^2x - cos^2xsinx/sin^2x correct?

Answer 19

Yep, see any other cancellations? c;

Answer 20

Woops, how'd u get a sin^2x on the bottom of your second fraction? :O

Answer 21

Remember, you were multiplying it by \(\large \dfrac{\sin x}{1}\)

Answer 22

(sorry, question here, are your little symbols just supposed to be lines and fancy x's?)

Answer 23

oh woops brain fart there. forgot that you dont multiply that on top and bottom - its over 1

Answer 24

Lines and fancy x's?? XD lol

The last fancy thing I sent should look like a fraction.
It's just to format the math so it's a lot easier to read.

Answer 25

ok so i cancelled and am now at sin^2x - cos^2x

Answer 26

haha that is what i mean - lol fractions ok because that is what im seeing so its all good

Answer 27

what do i do next?

Answer 28

We want to remember back to our most basic Trig Identity \(\large \sin^2x+\cos^2x=1\).
If we subtract cos^2x from each side, it gives us, \(\large \sin^2x=1-\cos^2x\).

It appears we're trying to get an answer that only involves cosine. So maybe we can use this identity. Hmmm?