# Trig identity questions

Hey guys,I need some help on the following trig identities:

1) sin2x = 2tanx/1+tan^2x

2) sin2x/sinx - cos2x/cosx = secx

My attempts:

1) LS: sin2x
2sinxcosx

2sinx/cosx

2tanx/1+tanx

Not sure if this is right or not. I kind of understand my third step but it just doesn't seem right, any validations? If not, help please.

2) LS: sin2x/sinx - cos2x/cosx

2sinxcosx/sinx - ???

sinx/cosx - ?? i have no idea where to go from here

verty
Homework Helper
Do #1 in reverse, it should be easier, and take little steps, don't make your steps too large. Each step should be small enough to be obvious to the reader.

HallsofIvy
Homework Helper
Hey guys,I need some help on the following trig identities:

1) sin2x = 2tanx/1+tan^2x

2) sin2x/sinx - cos2x/cosx = secx

My attempts:

1) LS: sin2x
2sinxcosx
Yes, sin(2x)= 2sin(x)cos(x)

2sinx/cosx
What? 2 sin(x)/cos(x)= 2 tan(x). Have you changed to the right side now??

2tanx/1+tanx
Where did this come from? It is what you want to arrive at but what are you doing to get it?

Not sure if this is right or not. I kind of understand my third step but it just doesn't seem right, any validations? If not, help please.

2) LS: sin2x/sinx - cos2x/cosx

2sinxcosx/sinx - ???

sinx/cosx - ?? i have no idea where to go from here

Frankly, I don't understand what you are trying to do. Please state exactly what you are doing to go from one step to another.

I just stated that I was trying to turn the left side into the right, sorry for the confusion. When I do these I just pick a side then use identities until it is equal to the opposite side.

tiny-tim
Homework Helper
hi tg22542!

(try using the X2 button just above the Reply box )
1) sin2x = 2tanx/1+tan^2x

2) sin2x/sinx - cos2x/cosx = secx

My attempts:

1) LS: sin2x
2sinxcosx

2sinx/cosx

2tanx/1+tanx

as verty says, it would be a lot easier to go from right to left!

if you must go from left to right, use cos = 1/sec
2) LS: sin2x/sinx - cos2x/cosx

first step is obviously to put them over a common denominator

okay so I believe I completed 1)

RS:

2tanx/1+tan^2x

2tanx/sec^2x

2sinx2cosx/sec^2x

2sinxcosx

sin2x

good? :)

thank you guys very much for your help so far, very appreciated

nevermind.. just realized the identity I used for sec is actually sec^2..

ugh

MarneMath
okay so I believe I completed 1)

RS:

2tanx/1+tan^2x

2tanx/sec^2x

2sinx2cosx/sec^2x

2sinxcosx

sin2x

good? :)

thank you guys very much for your help so far, very appreciated

Looks good to me.

For your other question, do you know any trig identities dealing with cos[2x]? Such as Cos[2x] = Cos[x]^2 - Sin[x]^2 ?

Mark44
Mentor
okay so I believe I completed 1)

RS:
The expressions below are all equal, so you should denote them as such by using =.

Also, what you wrote below needs parentheses around the terms in the denominator. What you wrote is really this:
$$\frac{2tan(x)}{1} + sec^2(x)$$
2tanx/1+tan^2x

2tanx/sec^2x

2sinx2cosx/sec^2x

2sinxcosx

sin2x

good? :)

thank you guys very much for your help so far, very appreciated

tiny-tim
Homework Helper
2tanx/1+tan^2x

2tanx/sec^2x

yes
2sinx2cosx/sec^2x

where did this come from?

use 1/sec = cos (as i said before)​

So could it go :

RS: 2tanx/1+tan^2x

=sin^2x/cos^2x + 1/cos^2x ??

tiny-tim
Homework Helper
(try using the X2 button just above the Reply box )
RS: 2tanx/1+tan^2x

=sin^2x/cos^2x + 1/cos^2x ??

i honestly have no idea how you got that

tg22542, in the exam, you must do one step at a time …

then the examiner can see what you're doing, and why, and at least give you some marks for getting something right!

try again, slowwwwly

RS: 2tanx/1+tan^2x

So I first used the identity tan^2x + 1 = sec^2x

So

= 2tanx + sec^2x

Next I used tanx=sinx/cosx

So:

2sinx/2cosx + sec^2x

Now I'm lost, It must equal 2sinx so I clearly have to get the cos to cancel out somehow, but I don't know which identities to use

tiny-tim
Homework Helper
RS: 2tanx/1+tan^2x

So I first used the identity tan^2x + 1 = sec^2x

correct

(except it would be clearer if you used brackets: 2tanx/(1+tan^2x) …

i suspect that's partly the reason why your next step was wrong …

as Mark44 has already said, always use brackets (parentheses))

So

= 2tanx + sec^2x

NO!

= 2tanx/sec2x …

carry on from there

Any idea on which identity to use from these point on? I'm lost :(

CAF123
Gold Member
Any idea on which identity to use from these point on? I'm lost :(

What is tanx equal to? What is sec2x equal to? Now sub in.

=(2sinx/2cosx)/(1/cos^2x) ?

tiny-tim
Homework Helper
=(2sinx/2cosx)/(1/cos^2x) ?

yes

(except for one silly mistake )

and then?​

I have no idea, I feel likei cou,d multiply the top by 2cos leaving me 2sinx2cosx/cos which wou,d give me the answer? Correct?

tiny-tim
Homework Helper
I have no idea, I feel likei cou,d multiply the top by 2cos leaving me 2sinx2cosx/cos which wou,d give me the answer? Correct?

tg22542, i don't understand your defeatism

you say "i have no idea", even though you obviously do have an idea (and it's the correct one)

yes, your 1/(1/something) can obviously be cancelled out

it's your "2"s that need a second look

MarneMath
(a/b)/(c/d) = (ad/bc) and then all you have to do is remember one more trig id. It would also be beneficial to not write 2tan(x) as (2sinx)/(2cosx), because that clearly isn't right.

I seriously don't know where to go from here..haha I'm just getting more and more confused

CAF123
Gold Member
I seriously don't know where to go from here..haha I'm just getting more and more confused

You are really there, all you need to correct is your statement that $$2\tan x = \frac{2 \sin x}{2 \cos x}$$ Correct this and you are effectively done.

Oh okay so

(2sinx/2cosx)/(1/cos^2x)

2cosx and cos^2x cancel leaving me with 2sinx

:) done

CAF123
Gold Member
Oh okay so

(2sinx/2cosx)/(1/cos^2x)

2cosx and cos^2x cancel leaving me with 2sinx

:) done

No. They don't cancel and, besides, what you want is sin2x, not 2sinx.

What is tanx equal to? Then what is 2 multiplied by tanx?