# Trig identity questions

1. Jun 10, 2013

### tg22542

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

2. Jun 10, 2013

### verty

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.

3. Jun 10, 2013

### HallsofIvy

Staff Emeritus
Yes, sin(2x)= 2sin(x)cos(x)

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

Where did this come from? It is what you want to arrive at but what are you doing to get it?

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.

4. Jun 10, 2013

### tg22542

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.

5. Jun 10, 2013

### tiny-tim

hi tg22542!

(try using the X2 button just above the Reply box )
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
first step is obviously to put them over a common denominator

6. Jun 10, 2013

### tg22542

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

7. Jun 10, 2013

### tg22542

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

ugh

8. Jun 10, 2013

### MarneMath

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 ?

9. Jun 10, 2013

### Staff: Mentor

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:
$$\frac{2tan(x)}{1} + sec^2(x)$$

10. Jun 11, 2013

### tiny-tim

yes
where did this come from?

use 1/sec = cos (as i said before)​

11. Jun 11, 2013

### tg22542

So could it go :

RS: 2tanx/1+tan^2x

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

12. Jun 11, 2013

### tiny-tim

(try using the X2 button just above the Reply box )
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

13. Jun 11, 2013

### tg22542

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

14. Jun 11, 2013

### tiny-tim

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))

NO!

= 2tanx/sec2x …

carry on from there

15. Jun 11, 2013

### tg22542

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

16. Jun 11, 2013

### CAF123

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

17. Jun 11, 2013

### tg22542

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

18. Jun 11, 2013

### tiny-tim

yes

(except for one silly mistake )

and then?​

19. Jun 11, 2013

### tg22542

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?

20. Jun 11, 2013

### tiny-tim

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