Results 1 to 2 of 2

Math Help - Need help with Trigonometric Identity questions.

  1. #1
    Newbie
    Joined
    Apr 2009
    Posts
    6

    Need help with Trigonometric Identity questions.

    Hey, need some help with two questions (may be adding more to this thread later). I am finding these problems rather difficult, so an explanation of each step needed would be EXTREMELY HELPFUL!
    1.  (secx - cosx)(cscx - sin x) = \frac{tanx}{1+tan^2x}
    My attempt:
    LHS
    =  (\frac{1}{cosx} - cosx)(\frac{1}{sinx} - sinx) *changed secx to 1/cosx, changed cscx to 1/sinx
    =  (\frac{sin^2x + cos^2x}{cosx} - cosx)(\frac{sin^2x + cos^2x}{sinx} - sinx) *replaced "1" with pythagrean
    This is where I am stuck as to what to do.

    2.  sec^6x - tan^6x = 1 + 3tan^2x sec^2x
    My attempt:
    LHS
    =  (\frac{1}{cos^6x}) - (\frac{sin^6x}{cos^6x}) *changed tanx to sinx/cosx, changed secx into 1/cosx
    =  (\frac {sin^2x + cos^2x} {cos^6x}) -  (\frac{sin^6x}{cos^6x}) * Replaced "1" with pythagrean
    Unknown what to do from here.





    EDIT: Okay, I just read the rules. My apologies for all with the title involving "help" I should have been more specific. Unknown if possible to change title or not, I do not see a way right now.
    What I did in the edit:
    Showed my attempts and tried to explain what I was thinking for each step.

    EDIT2: K, everything is in LaTeX! Took about 10 minutes extra, but WOW, looks a lot neater and easier to understand! Really glad I read the rules.
    Last edited by Xenophobe; April 18th 2009 at 08:09 PM.
    Follow Math Help Forum on Facebook and Google+

  2. #2
    MHF Contributor Reckoner's Avatar
    Joined
    May 2008
    From
    Baltimore, MD (USA)
    Posts
    1,024
    Thanks
    75
    Awards
    1

    Smile

    I did these through sheer brute force. You may be able to clean it up a bit.

    (\sec x-\cos x)(\csc x-\sin x)=\frac{\tan x}{1+\tan^2x}
    Expand the left side:

    (\sec x-\cos x)(\csc x-\sin x)

    =\sec x\csc x-\sin x\sec x-\cos x\csc x+\sin x\cos x

    =\sec x\csc x-\tan x-\cot x+\sin x\cos x

    =\frac1{\sin x\cos x}+\sin x\cos x-\tan x-\cot x

    =\frac{1+\sin^2x\cos^2x-\sin^2x-\cos^2x}{\sin x\cos x}

    =\frac{1+\sin^2x\cos^2x-1}{\sin x\cos x}=\frac{\sin^2x\cos^2x}{\sin x\cos x}

    =\sin x\cos x

    =\frac{\sin x}{\sec x}=\frac{\sin x\sec x}{\sec^2x}

    and you can hopefully finish this last step.

    \sec^6x-\tan^6x=1+3\tan^2x\sec^2x
    \sec^6x-\tan^6x

    =\frac{1-\sin^6x}{\cos^6x}

    =\frac{\left(1-\sin^2x\right)\!\left(1+\sin^2x+\sin^4x\right)}{\c  os^6x}

    =\frac{\cos^2x\left(1+\sin^2x+\sin^4x\right)}{\cos  ^6x}

    =\frac{1+\sin^2x+\sin^4x}{\cos^4x}

    =\sec^4x+\tan^2x\sec^2x+\tan^4x

    =\sec^4x+\left(\sec^2x-1\right)\sec^2x+\tan^4x

    =2\sec^4x-\sec^2x+\tan^4x

    =2\sec^4x-\sec^2x+\left(\sec^2x-1\right)^2

    =3\sec^4x-3\sec^2x+1

    =3\sec^2x\left(\sec^2x-1\right)+1

    =1+3\tan^2x\sec^2x
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. [SOLVED] trigonometric sum identity
    Posted in the Differential Geometry Forum
    Replies: 11
    Last Post: September 29th 2010, 08:06 PM
  2. Trigonometric identity - HELP please
    Posted in the Trigonometry Forum
    Replies: 2
    Last Post: June 17th 2010, 10:12 PM
  3. Trigonometric identity help
    Posted in the Trigonometry Forum
    Replies: 4
    Last Post: April 1st 2010, 03:35 PM
  4. trigonometric identity
    Posted in the Trigonometry Forum
    Replies: 1
    Last Post: May 3rd 2009, 06:35 PM
  5. trigonometric identity
    Posted in the Trigonometry Forum
    Replies: 5
    Last Post: May 5th 2008, 04:42 AM

Search Tags


/mathhelpforum @mathhelpforum