Trig identity questions

  • Thread starter tg22542
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  • #1
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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
 

Answers and Replies

  • #2
verty
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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
HallsofIvy
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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.
 
  • #4
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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
tiny-tim
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hi tg22542! :smile:

(try using the X2 button just above the Reply box :wink:)
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 :smile: 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 :wink:
2) LS: sin2x/sinx - cos2x/cosx

first step is obviously to put them over a common denominator
 
  • #6
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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
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nevermind.. just realized the identity I used for sec is actually sec^2..

ugh
 
  • #8
MarneMath
Education Advisor
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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 ?
 
  • #9
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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
 
  • #10
tiny-tim
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2tanx/1+tan^2x

2tanx/sec^2x

yes :smile:
2sinx2cosx/sec^2x

where did this come from? :confused:

use 1/sec = cos (as i said before)​
 
  • #11
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So could it go :

RS: 2tanx/1+tan^2x

=sin^2x/cos^2x + 1/cos^2x ??
 
  • #12
tiny-tim
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(try using the X2 button just above the Reply box :wink:)
RS: 2tanx/1+tan^2x

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

i honestly have no idea how you got that :confused:

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 :smile:
 
  • #13
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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
tiny-tim
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RS: 2tanx/1+tan^2x

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

correct :smile:

(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 :wink:
 
  • #15
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Any idea on which identity to use from these point on? I'm lost :(
 
  • #16
CAF123
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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.
 
  • #17
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=(2sinx/2cosx)/(1/cos^2x) ?
 
  • #18
tiny-tim
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=(2sinx/2cosx)/(1/cos^2x) ?

yes :smile:

(except for one silly mistake :wink:)

and then?​
 
  • #19
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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
tiny-tim
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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) :confused:

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

it's your "2"s that need a second look :redface:
 
  • #21
MarneMath
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(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.
 
  • #22
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I seriously don't know where to go from here..haha I'm just getting more and more confused
 
  • #23
CAF123
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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.
 
  • #24
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Oh okay so

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

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

:) done
 
  • #25
CAF123
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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?
 

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