Results 1 to 5 of 5

Math Help - Identities

  1. #1
    Newbie
    Joined
    Oct 2006
    Posts
    4

    Smile Identities

    Could anybody give me some help with these identities that I need to complete for homework? I need to have them correct with all work for tomorrow, and they need to be proofs. If I could at least get help with starting them, that would be great. Thanks.

    1) (csc X - sin X)/(cot^2 X) = sin X

    2) (tan X - sin X)/(sin^3 X) = (sec X)/(1+cos X)

    3) ((cos X)/(1- sin X)+(1+ sin X)/(cos X)) = 2(sec X + tan X)


    Again, any help will be appreciated.
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Global Moderator

    Joined
    Nov 2005
    From
    New York City
    Posts
    10,616
    Thanks
    10
    Quote Originally Posted by anna_sims View Post

    1) (csc X - sin X)/(cot^2 X) = sin X
    Express everything in terms of sine and cosine.

    \frac{\frac{1}{\sin x}-\sin x}{\frac{\cos^2 x}{\sin^2 x}}
    Add the numerator,
    \frac{\frac{1-\sin^2 x}{\sin x}}{\frac{\cos^2 x}{\sin^2 x}}
    Pythagoren identities,
    \frac{\frac{\cos^2 x}{\sin x}}{\frac{\cos^2 x}{\sin^2 x}}
    Fractions means divison thus,
    \frac{\cos^2 x}{\sin x}\div \frac{\cos^2 x}{\sin^2 x}
    Thus,
    \frac{\cos^2 x}{\sin x}\cdot \frac{\sin^2 x}{\cos^2 x}
    Cancel,
    \frac{\sin x}{1}=\sin x
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Super Member

    Joined
    May 2006
    From
    Lexington, MA (USA)
    Posts
    11,914
    Thanks
    778
    Hello, Anna!

    3)\;\frac{\cos x}{1- \sin x} + \frac{1 + \sin x}{\cos x} \:=\:2(\sec x + \tan x)

    Multiply the first fraction by \frac{1 + \sin x}{1 + \sin x}

    . . \frac{\cos x}{1 - \sin x}\cdot\frac{1 + \sin x}{1 + \sin x} \:=\:\frac{\cos x(1 + \sin x)}{1 - \sin^2x} \:=\:\frac{\cos x(1 + \sin x)}{\cos^2x} \:=\:\frac{1 + \sin x}{\cos x}


    The problem becomes: . \frac{1 + \sin x}{\cos x} + \frac{1 + \sin x}{\cos x} \:=\:2\left(\frac{1 + \sin x}{\cos x}\right)

    . . = \:2\left(\frac{1}{\cos x} + \frac{\sin x}{\cos x}\right) \:=\:2(\sec x + \tan x)

    Follow Math Help Forum on Facebook and Google+

  4. #4
    Super Member
    earboth's Avatar
    Joined
    Jan 2006
    From
    Germany
    Posts
    5,830
    Thanks
    123
    Quote Originally Posted by anna_sims View Post
    Could anybody give me some help with ...
    2) (tan X - sin X)/(sin^3 X) = (sec X)/(1+cos X)
    ...
    Hi,

    I'll take the RHS of the equation and show that it is equal to the LHS:

    \frac{\tan(x)-\sin(x)}{\sin^3(x)}=\frac{\sec(x)}{1+\cos(x)}
    =\frac{\frac{1}{\cos(x)}}{1+\cos(x)}\cdot \frac{1-\cos(x)}{1-\cos(x)}
    =\frac{\frac{1}{\cos(x)}-1}{1-\cos^2(x)}\cdot \frac{\sin(x)}{\sin(x)}
    =\frac{\frac{\sin(x)}{\cos(x)}-\sin(x)}{\sin^3(x)}
    =\frac{\tan(x)-\sin(x)}{\sin^3(x)}

    EB
    Follow Math Help Forum on Facebook and Google+

  5. #5
    Super Member

    Joined
    May 2006
    From
    Lexington, MA (USA)
    Posts
    11,914
    Thanks
    778
    Nice job, Earboth!

    I tried going from left to right . . . not worth the effort!


    2)\;\frac{\tan x - \sin x}{\sin^3x} \:= \:\frac{\sec x}{1 + \cos x}

    We have: . \frac{\frac{\sin x}{\cos x} - \sin x}{\sin^3x}\;=\;\frac{\sin x\left(\frac{1}{\cos x} - 1\right)} {\sin^3x}


    Multiply by \frac{\cos x}{\cos x}:\;\;\frac{\cos x}{\cos x} \cdot \frac{\sin x\left(\frac{1}{\cos x} - 1\right)}{\sin^3x}\;=\;\frac{\sin x(1 - \cos x)}{\sin^3x\cos x} \;=\;\frac{1-\cos x}{\sin^2x\cos x}

    . . = \;\frac{1-\cos x}{(1-\cos^2x)\cos x} \;=\;\frac{1-\cos x}{(1-\cos x)(1 + \cos x)\cos x}\;=\;\frac{1}{(1+\cos x)\cos x}

    Therefore: . \frac{1}{1+\cos x}\cdot\frac{1}{\cos x} \;=\;\frac{\sec x}{1 + \cos x}

    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Replies: 6
    Last Post: June 23rd 2010, 12:59 AM
  2. Identities
    Posted in the Pre-Calculus Forum
    Replies: 1
    Last Post: March 7th 2010, 11:19 PM
  3. Identities
    Posted in the Trigonometry Forum
    Replies: 4
    Last Post: March 7th 2010, 06:39 PM
  4. set identities
    Posted in the Discrete Math Forum
    Replies: 5
    Last Post: April 12th 2009, 03:57 PM
  5. Identities help...
    Posted in the Trigonometry Forum
    Replies: 1
    Last Post: November 13th 2008, 08:30 AM

Search Tags


/mathhelpforum @mathhelpforum