How to Solve Trig Identity Questions: Tips and Examples

[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

 

[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.

 

[HallsofIvy]

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.

 

[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.

 

[tiny-tim]

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

 

[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

 

[tg22542]

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]

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]

2tanx/1+tan^2x

2tanx/sec^2x

yes

2sinx2cosx/sec^2x

where did this come from?

use 1/sec = cos (as i said before)​

 

[tg22542]

So could it go :

RS: 2tanx/1+tan^2x

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

 

[tiny-tim]

(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 ​

 

[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

 

[tiny-tim]

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 ​

 

[tg22542]

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

 

[CAF123]

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.

 

[tg22542]

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

 

[tiny-tim]

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

yes

(except for one silly mistake )

and then?​

 

[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?

 

[tiny-tim]

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 canceled 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.

 

[tg22542]

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

 

[CAF123]

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.

 

[tg22542]

Oh okay so

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

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

:) done

 

[CAF123]

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?

 

[tg22542]

2sinx/2cosx?

 

[CAF123]

2sinx/2cosx?

No. You figured that $$\tan x = \frac{\sin x}{\cos x}$$ What happens when you multiply this by 2?

You should know that $$\frac{a}{b} \times \frac{c}{d} = \frac{ac}{bd}$$

 

[MarneMath]

Just to give you a concrete example. If you have 2*(4/2) that wouldn't be the same as (2*4)/(2*2), so why do you think 2(sinx/cosx) is the same as (2sinx)/(2cosx)?

 

[tiny-tim]

2sinx/2cosx?

tg22542, how would you simplify 2sinx/2cosx?