1. Not finding help here? Sign up for a free 30min tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Indentites - Calculus (newbie)

  1. Apr 21, 2004 #1
    Hi all.
    Hey - i'm new here - so please excuse me if i have posted in the wrong thread etc
    Now, for the details.
    Firstly, I'm from New Zealand, 16yrs - at High School.
    I'm doing NCEA (new system) level 3 Calculus (last year High School Calc)
    I came here mainly to look around - but i have a problem, so i thought I might as well ask.

    We're doing Trigonometry at the moment - and we're touching on Indentites - and proving them.
    Me and my mate have been going through doing the worksheet we got given for the holidays - we've managed them all ok, except from one, so i thought i'd come hear for help (like i said earlier - this may be the wrong thread - o, and what is "college" and Grades K-12 - what would i fit into...)

    I'm not to sure if this is standard stuff or not... but anyway, here goes...

    We have to prove that:

    cos²x cosec²x - sin²x sec²x = 4 cot2x cosec2x

    Now, like I said - i tried it by myself - no luck, went to my classmate - we spent about 3 hrs straight, and 20-30 pages, and were completey puzzled.

    If I do it in the simplistic way & change the Left side, I change:
    cosec²x - which is 1/sin²x (correct?)

    and sec²x to 1/cos²x (correct?)
    mulitply cos²x with sin²x = cos²x/sin² x
    and the other one = sin²x/cos²x
    Subtract the two, and i get = 1 -- which is not true...

    Another method we tried was...
    as cosec²x = cot²x + 1
    and
    sec²x = tan²x + 1
    you can subsitute them in, then change the cot²x to cos²x/sin²x (is that true? will they both be squared?)
    and the tan²x to sin²x/cos²x

    but even with this way we ended up in a jumble mess

    We also tried starting by changing the right hand side (the 4 cot2x cosec2x) - even though we were talk not to use this method UNLESS YOU HAD TO, but we still got no where.

    There were heaps of other methods we used - but I'd be here all night typing them up.
    So if anyone has any ideas - i really only needed a pointer, or hint - i'll be able to sus out the rest! and I don't like 'cheating'

    Thanks
    Alex
     
  2. jcsd
  3. Apr 21, 2004 #2
    I don't know if there are shorter ways of doing it, but the whole solution took me one page, and I have really small handwriting. Anyway, here's where you went wrong:

    Check it again. When you subtract the two, it does not result in a one. Correct this and continuing solving it using this method (not the second one).
     
  4. Apr 21, 2004 #3
    It also helps to change the Right-hand Side. In order to change the RHS, it will be useful for you to know the double-angle formulas. Do you know them already?
     
  5. Apr 21, 2004 #4

    HallsofIvy

    User Avatar
    Staff Emeritus
    Science Advisor

    For most people, it is simplest to convert everything to sine and cosine- since we are more used to them. In this example, cos2(x)cosec2(x) becomes cos2(x)/sin2(x) and sin2(x)sec(x) becomes
    sin2(x)/cos2(x). Now get common denominators and subtract the fractions:
    [tex]\frac{cos^2(x)}{sin^2(x)}- \frac{sin^2(x)}{cos^2(x)}= \frac{cos^4(x)- sin^4(x)}{sin^2(x)cos^2(x)}[/tex]

    You should be able to recognize cos4(x)- sin4(x) as a "difference of squares": it is equal to (cos2(x)- sin2(x)(cos2(x)+ sin2(x))- and we all know what cos2(x)+ sin2(x) is!

    The left-hand side reduces to
    [tex]\frac{cos^2(x)- sin^2(x)}{sin^2(x)cos^2(x)}[/tex]

    I seem to recall an identity that says "cos2(x)- sin2(x)= cos(2x)" as well as "2sin(x)cos(x)= sin(2x)". Since the right-hand side has "2x", those should be useful! (Aha! Now I see where that "4" on the right-hand side came from!)
     
  6. Apr 22, 2004 #5
    Hi.
    Hey - thanks guys for that.
    I went bak thro all our working - dunno why i thought that equalled one...
    I see i did it like that (=1) on the very first attempt - and when i went through with my mate - we only spent a few mins on it (doing it that way) and we got one aswell - our working was correct up to the subtraction part - i guess we just had a block or something.

    Anyway - thanks - i thought it was something simple - or something to hard that we hadn't been taught.
    I'll go work on it, and get back to you
     
  7. Apr 22, 2004 #6
    Hmmmm.
    Ok, i'm a little confused.
    We only really touched on Double Angles on the last 2 days of the term (and you know what the last few days of term are like...)
    Yeah, i have all those identities of them - so thats ok.
    Going 'HallsofIvy' way - i also get it down to:

    cos²(x)-sin²(x)
    --------------
    sin²(x)cos²(x)

    As cos2(x)+ sin2(x) = 1

    Thats all cool, but i trip over at the next stage.
    Yup, you're right on "cos2(x)- sin2(x)= cos(2x)" - but i also have noted that it can also become =1-2sin²(x)

    But what you do at the next stage I don't understand...
    >as well as "2sin(x)cos(x)= sin(2x)"

    What is that referring to?

    I get to:
    Cos(2x) or 1-2sin²(x)
    ----------- -----------
    sin²(x)cos²(x) sin²(x)cos²(x)

    hmmmm...
    nope, I'm lost :(

    And 'recon' - i have a few Q's about Double Angles and changing the right hand side:
    1) when you change 4cot(2x) - what does it work out to - (what happens to the 2x)
    Does it become : 4cos(2x)
    --------
    sin (2x) or is the 2x only on one side? and if so, which, and why?

    So if my above method is correct - i get this:
    4cos(2x)
    --------
    sin²(4x)

    And then - changing 4cos(2x) - using the unfamilar Double Angles -
    2cos²(x)-2sin²(x)

    na, i'm sure i'm just drifting off here - i should have taken more notice in that last class of Calc!

    But going backwards - I some how have to get sin²(x)cos²(x) to = 4/sin (2x) then i'm sweet!
    But thats the link I'm missing...
     
  8. Apr 22, 2004 #7
    tan(2x) = [1 - tan^2(x)]/2 tan (x)

    Since cot(x) = 1/tan(x) so

    cot(2x) = 1/tan (2x) = 2 tan (x)/[1 - tan^2(x)]
     
  9. Apr 22, 2004 #8
    I did not use the double angle formula for cos (2x) in solving the problem.
     
  10. Apr 25, 2004 #9
    Ok, i'm still not thinking on the right lines here....

    I can get to this:

    cos(2x)
    -----------
    sin²(x)cos²(x)

    on left side
    and this:

    4/tan(2x) . 1/sing(2x) on right

    And changing the Tan like Recon said...

    4(1-tan²(x) 1
    ----------- . ----
    2tan x sin(2x)

    but thats just going to end up in a messs.


    I'm sure I'm missing something on the left side.

    Recon - did you get to where I am now? If not - could you tell me which side i have went rong on, and how to fix it up.
    And did you alter both sides so they become equal - but not showing what the question asked (i hope thats worded right...!)
     
  11. Apr 25, 2004 #10
    'HallsofIvy' has already solved your problem Sir

    once you get cos(2x) in the numerator and sin(2x)/2 squared in the denomenator, the problem is solved!!!

    Look how,

    cos(2x)/(sin(2x)/2) = 2cot(2x) right!
    and 1/(sin(2x)/2) = 2cosec(2x) OK!

    now multiply the two and you get 4cot(2x)cosec(2x)
     
  12. Apr 25, 2004 #11
    Fergie, forget about my way of solving this problem and stick with HallsofIvy. His method is easier to understand. Mine goes on for many many lines. Anyway, I thought I might outline HallsofIvy and Xishan's solution more clearly:

    [tex]2sin(x)cos(x) = sin(2x)[/tex] --> Double angle formula

    So [tex]sin(x)cos(x) = \frac{sin(2x)}{2}[/tex]
    And squaring both sides,

    [tex]sin^2(x)cos^2(x) = \frac{sin^2(2x)}{4}[/tex] - Just plug this in the equation below.

    As others have pointed out [tex]cos^2(x)- sin^2(x)= cos(2x)[/tex]

    So after doing some simple algebra you get

    [tex]\frac{4cos(2x)}{sin^2(2x)}[/tex]

    Then have a look at Xishan's post.
     
    Last edited: Apr 25, 2004
  13. Apr 25, 2004 #12
    Thanks for further clarifying my point!
     
  14. Apr 27, 2004 #13
    hi.
    thanks all!
    I think i have it sorted.
    You guys cleared a few things up for me - thanks a heap.

    No doubt i'll be back again in the future
    Thanks
    Alex
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?



Similar Discussions: Indentites - Calculus (newbie)
Loading...