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More Trig Identity Proofs .

  • Thread starter VanKwisH
  • Start date
  • #1
108
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More Trig Identity Proofs .........

Homework Statement


1. cot^2x - 1 = cot2x
-----------
2cotx


2. tanx + cotx = 2csc2x


3. cos(A+B) = 1-tanAtanB
--------- ----------
cos(A-B) 1+tanAtanB


Homework Equations





The Attempt at a Solution



Anyone how to do these proofs??
i've tried simplifying all of these as much as i could but i don't seem to get the
right answer ...... can anyone show me???
 
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Answers and Replies

  • #2
662
0
What have you tried? Show us your attempts, and we'll guide you toward the answer.

You might start just by writing the identities that you think might be relevant.
 
  • #3
108
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okay lets work on the first one then ..............

i'll work on the left hand side.........

on the top i know that cot^2x = 1 / tan^2x = cos^2x / sin^2x
and i know that 2cotx is 2 * cosx / sinx

that is all i have so far ....... what next?? use complex fraction ?? or find common denominator??
 
  • #4
662
0
Okay, you need some common identities. cos^2x and sin^2x have a simple relation so that you never need to have both of them - in other words you can always get rid of one and be left with only the other.

Also, here's a key to problems of this kind: you see that the expressions on one side have functions of x whereas the expressions on the other side have functions of 2x? How you you relate those? You're going to have to be able to get from functions of 2x to functions of x - what identities tie those together?
 
  • #5
108
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ahhhhhhh but there's so many differernt expresions ..... which one should i use ???
 
  • #6
108
0
i see that cot2x has the function of 2x .......... so that would make is cos2x / sin2x ..
then cos2x = cos^2x - sin^2x
and sin2x = 2sinxcosx so it's now
cos^2x - sin^2x
-----------------
2sinxcosx
 
  • #7
108
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now what about the left side??
cot^2x - 1
-----------
2cotx
 
  • #8
108
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OOOOOOOOO NVM I GOT IT !!!!!!!!!!!!!!!
how about number 2??

2. tanx + cotx = 2csc2x
 
  • #9
662
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<snip> ...
cos2x = cos^2x - sin^2x
and sin2x = 2sinxcosx ...
Those are the ones! Almost everything can be expressed in terms of sines and cosines, so those two
are your best friends for the rest of the evening.
 
  • #10
662
0
OOOOOOOOO NVM I GOT IT !!!!!!!!!!!!!!!
how about number 2??

2. tanx + cotx = 2csc2x
Okay, I'll tell you - I have no idea off the top of my head, but I'll bet $5 that if you express everything in terms of sines and cosines you'll be half the way there ....
 
  • #11
108
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got it .................... in sines and cosines
left side
( sinx / cosx ) + ( cosx / sinx )

right side

( 2 / sin2x )

that's as far as i got ....... but what should i do now ???
can i find a common denominator for the left side???
and also can i simplify the right side even more??
 
  • #12
rock.freak667
Homework Helper
6,230
31
When proving Trig Identities, work with one side only. Bring the left side to the same denominator. Then apply some more trig identities that you know.
 
  • #13
662
0
Keep going - you've got it. Just simplify- I'd start with the left side and then see if you can get the right side to look the same.
 
  • #14
662
0
When proving Trig Identities, work with one side only. Bring the left side to the same denominator. Then apply some more trig identities that you know.
That's true - the procedure is generally to rearrange one side so that it becomes the same as the other. Sometimes, though, it helps to work "backwards" from the other side a little, and then when you've got the same thing on both sides, to reverse those steps so that you do all the steps on one side.
 
  • #15
108
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okay on the left side I now have
sin^2x + cos^2x ....
it can either be
1 .... which is clearly wrong ....
or 1-cos^2x + 1 - sin^2x
Can i re-write this into something else??

or can it be something else which i have no idea...
 
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  • #16
662
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Okay, I'm not sure how you've gotten where you are. You had:
( sinx / cosx ) + ( cosx / sinx )

What do you get when you add these two? It should be a pretty simple expression.

Now think about that right hand side - how can you get to that expression? (I'd work it backwards and then reverse your steps to complete the calculation on the left hand side.)
 
  • #17
108
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OOO when i add i get
( sinx + cosx / sinx*cosx ) am i right????
 
  • #18
1,752
1
These are the main ones you need to know ...

http://alt2.mathlinks.ro/Forum/latexrender/pictures/e/0/f/e0f3f955ccf55a86c49f7fa5d4e0454061f7cbdc.gif [Broken]
 
Last edited by a moderator:
  • #19
662
0
OOO when i add i get
( sinx + cosx / sinx*cosx ) am i right????
Nope!! Be careful - you had to multiply numerators and denominators by sinx (for the first one) or cosx, right?
 
  • #20
108
0
Nope!! Be careful - you had to multiply numerators and denominators by sinx (for the first one) or cosx, right?
i thought i had to find a common denominator .............. and that would be sinxcosx...
 
  • #21
108
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yargh can u just show me the solution already ...PLEASE!!!... this is really stressing me out waiting for your reply ..... and i have lots of other homework i have to attend to ......
it would be greatly appreciated
 
  • #22
1,752
1
yargh can u just show me the solution already ...PLEASE!!!... this is really stressing me out waiting for your reply ..... and i have lots of other homework i have to attend to ......
it would be greatly appreciated
Keep working at it. Once you get through these concepts your hw will be a breeze.
 
  • #23
662
0
Sorry ... I'm bouncing between different threads.

You did get a common denominator, but you forgot to multiply the numerators by the sinx and cosx factors that you needed to get the denominators the same. Doing that would give you squares of sinx and cosx on top.

Now go to the RHS and use the identity for sin(a+b) to expand csc(2x) (write it as 1/sin first). Both sides should now look the same.
 
  • #24
108
0
Sorry ... I'm bouncing between different threads.

You did get a common denominator, but you forgot to multiply the numerators by the sinx and cosx factors that you needed to get the denominators the same. Doing that would give you squares of sinx and cosx on top.

Now go to the RHS and use the identity for sin(a+b) to expand csc(2x) (write it as 1/sin first). Both sides should now look the same.
alright my left hand side is .....
(sin^2x*cosx / cosxsinx ) + (sinx*cos^2x / sinxcosx )

what do u mean by
"Now go to the RHS and use the identity for sin(a+b) to expand csc(2x) (write it as 1/sin first)"

my right hand side is
2 * 1/sin2x = 2/sin2x

is that what u mean??
 
  • #25
662
0
You've got the LHS wrong. You might have to relax and just slow down a bit.

You should have multiplied sinx/cosx by sinx/sinx, and cosx/sinx by cosx/cosx. That should have given you sin^2x + cos^2x on the top.

On the RHS you have a sin(2x). What's that equal to? Remember, you have identities for relating sinx to sin(2x)=sin(x+x).
 

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