Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Homework Help: Need help with some identity problems

  1. Jun 29, 2008 #1
    I'm having difficulties with a few identity problems and I wanted to make sure I'm doing the ones I believe I did correctly, correctly...

    1. (cos^3x)+(sin^2x)(cosx)
    (cosx)(cos^2x)+(sin^2x)(cosx)
    2cosx

    2. (1+cosy)/(1+secy)
    (1+cosy)/(1+1/cosy)
    (1+cosy)/(1+cosy)
    1

    3. (tanx)/(secx)
    (sinx/cosx)/(1/cosx)
    (sinx/cosx)(cosx)
    sinx

    4.(secx-cosx)/(tanx)
    (secx-cosx)(cosx/sinx)
    (1/sinx)-(cos^2x/sinx)
    (1-cos^2x)/(sinx)
    (1/sinx)
    cscx

    5.(sinx/cscx)+(cosx/secx)
    (sinx/1/sinx)+(cosx/1/cosx)
    (sinxsinx)+(cosxcosx)
    sin^2x+cos^2x
    1

    6. (1+sinx/cosx)+(cosx/1+sinx)
    (1+sinx/1+sinx0(1+sinx/cosx)+(cosx/cosx)(cosx/1+sinx)
    (1+2sinx+sin^2x+cos^2x)/(1+sinx)(cosx)
    (2+2sinx)/(1+sinx)(cosx)
    (2+sinx)/(cosx)
    2tanx

    7. (tanx)(cosx)(cscx)
    (sinx/cosx)(cosx)(1/sinx)
    1

    The ones that I'm not too sure about are 2 and 8. I feel pretty confident about the other ones, but I wanted to make sure I'm doing them correctly.

    Thanks a lot to anybody that helps.
     
  2. jcsd
  3. Jun 29, 2008 #2

    rock.freak667

    User Avatar
    Homework Helper

    For 2
    From the text in red, multiply both the numerator and denominator by cosy


    I don't see a question 8...the only one that looked wrong at a glance was 2.
     
  4. Jun 29, 2008 #3
    First, you did not put a #8 up so I guess you are confident about that one lol.

    I didn't look at #6 b/c my eyes are tired, but you will want to check 2 and 4!

    for #2 you have that 1+(1/cos(y)) = 1 + cos(y) !! since when is that true. instead, you should multiply the numerator and denominator by cos(y) then see what you get.

    #4 the hint is to look at what 1 - (cos(x))^2 is
     
  5. Jun 29, 2008 #4

    HallsofIvy

    User Avatar
    Science Advisor

    Where did you get the "2"?

    (cosx)(cos^2x)+ (sin^2x)(cosx)= cos(x)(cos^2x+ sin^2x).

    How did you make this jump? 1/cosy is NOT equal to cos y.

    Yes, that's right.

    You've got your parenthes wrong. Also sec x= 1/cosx not 1/sin x
    (1/cosx- cosx)(cos(x)/sin(x)= (1/cox)(cosx/sinx)- cos^2x/sinx= 1/sinx- cos^2x/sinx
    = (1- cos^2x)/sinx

    Yes, that's correct.

    Again, you are not being careful with parentheses. I think you mean
    (1+ sinx)/cosx+ cosx/(1+ sin(x)) bit what you wrote is 1+ (sinx/cosx)+ cosx+ sin(x).

     
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook