Question

i have no idea ho to do this
prove the trig identity

1-cos(2x)/tan(x) =sin(2x)

Answer 1

\[1-\cos (2x)/\tan(x)=\sin(2x)\]

Answer 2

What you want to do is to somehow turn the left-hand side of the equation into the right-hand side...

Answer 3

@terenzreignz stilll dont get it :(

Answer 4

Let's first remove any possible ambiguity
\[\Large \frac{1-\cos(2x)}{\tan(x)}=\sin(2x)\]
Is it this?

Answer 5

yes @terenzreignz

Answer 6

Well then, let's see if we can use a little magic... ^_^
We have an established identity for
\[\large \cos(2x) =\color{red}?\]
What is it?

Answer 7

cos^2x-sin^2x

Answer 8

\[\cos^2x-\sin^2x\] @terenzreignz

Answer 9

Okay, let's go with that... so the left-hand side (specifically, the red part) becomes...
\[\Large \frac{1-\color{red}{\cos(2x)}}{\tan(x)}=\frac{1-\color{red}{[\cos^2(x)-\sin^2(x)]}}{\tan(x)}\]

Okay, catch me so far?

Answer 10

@terenzreignz a little yes..

Answer 11

Can't be a little, I need you to understand fully before we can proceed... is there something that is unclear to you?

Answer 12

why did u change sin(2x) to that
i thought it would be
\[2sinx*cosx\]

Answer 13

It was a cos(2x) remember? Not a sin(2x)
We don't care about the sin(2x) yet,
our "mission" is to turn the left-hand side of the equation into sin(2x)
Which we will... in time... so can we proceed?

Answer 14

@terenzreignz okok yesss

Answer 15

Okay great, now we simplify...
\[\Large \frac{1-\cos^2(x)+\sin^2(x)}{\tan(x)}\]

Everything okay so far?

Answer 16

@terenzreignz what operation did u do? subtraction, division, addition?

Answer 17

I just distributed the minus sign... remember, before that step, it looked like this...\[\Large \frac{1-\color{red}{[\cos^2(x)-\sin^2(x)]}}{\tan(x)}\]

Answer 18

@terenzreignz okokok! i see

Answer 19

Now this part here...
\[\Large \frac{\color{red}{1-\cos^2(x)}+\sin^2(x)}{\tan(x)}\]

Does it look familiar to you?

Answer 20

@terenzreignz yup it doessss!

Answer 21

then it simplifies into...?

Answer 22

Do remember the Pythagorean identity...
\[\Large \color{blue}{\sin^2(x)}+\color{red}{\cos^2(x)}= \color{red}1\]

Answer 23

@terenzreignz NOO :(

Answer 24

then you must... You can't go about proving identities without a thorough knowledge of the basic ones :)

Answer 25

in my notes it says
\[\sin ^2x+\cos ^2x=1\]
\[1+\tan ^2x=\sec ^2x\]
\[1+\cot ^2x=\csc ^2x\]

Answer 26

@terenzreignz

Answer 27

Yeah, those... the first one, is it not what I just wrote? :D

Answer 28

@terenzreignz ur rightttttt

Answer 29

So, given that first equation, if we subtract \(\cos^2(x)\) from both sides, you get...
\[\Large \sin^2(x) =\color{red}{1-\cos^2(x)}\]

Answer 30

@terenzreignz okok!! yesss

Answer 31

Well then, this part becomes...
\[\Large \frac{\color{red}{1-\cos^2(x)}+\sin^2(x)}{\tan(x)}\]

Answer 32

@terenzreignz OK... so now we got rid of the otherside

Answer 33

No, first, that part in red, it's just equal to
\[\Large \frac{\color{red}{\sin^2(x)}+\sin^2(x)}{\tan(x)}\]
right?

Answer 34

@terenzreignz yess righttt

Answer 35

So let's just combine them..
\[\Large \frac{\color{red}{2\sin^2(x)}}{\tan(x)}\]

Answer 36

@terenzreignz okok!! is there more ways to simplify it?

Answer 37

Of course...
\[\large \tan(x) =\frac{\sin(x)}{\cos(x)}\]

Answer 38

So it becomes...
\[\Large \frac{2\sin^2(x)}{\frac{\sin(x)}{\cos(x)}}\]

Answer 39

@terenzreignz ok i get what u did.

Answer 40

So we can bring the cos(x) up...
\[\Large = \frac{2\sin^2(x)\cos(x)}{\sin(x)}\]

Answer 41

And finally, cancel out the sin(x)
\[\Large = \frac{2\sin^{\cancel2}(x)\cos(x)}{\cancel{\sin(x)}}\]

Answer 42

Leaving you with...
\[\Large 2\sin(x)\cos(x)\]
Which, obviously, is...?

Answer 43

1? @terenzreignz

Answer 44

What? NO!
\[\Large \sin(2x) = 2\sin(x)\cos(x) \]

Answer 45

LOL sowwy :(

Answer 46

And that's it... we've successfully turned the left-side into the right side sin(2x)

Answer 47

@terenzreignz wowww thank u i learned alot!!!!:)

Answer 48

That's all i need to know :)