Results 1 to 7 of 7

Math Help - Simplifying trigonometric functions

  1. #1
    Member
    Joined
    Aug 2007
    Posts
    182

    Simplifying trigonometric functions

    Hi! I have started on several of these problems, but I am having a really hard time with them...could anyone please look over my work?
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Bar0n janvdl's Avatar
    Joined
    Apr 2007
    From
    South Africa
    Posts
    1,630
    Thanks
    6
    Quote Originally Posted by aikenfan View Post
    Hi! I have started on several of these problems, but I am having a really hard time with them...could anyone please look over my work?
    Number 5 seems fine.

    ---

    For Number 6:

    This one really isn't hard.

    Let's set cos^2 x = y

    y - sin^2 x = 2y - 1

    y - 2y - sin^2 x = -1

    -y - sin^2 x = - 1

    [Multiply by -1]

    y + sin^2 x = 1

    But \ y = cos^2 x

    cos^2 x + sin^2 x = 1


    ----

    Your solution to number 7 was a bit elaborate.

    Simply say:

     tan^2 x + 5 - 4 = sec^2 x

     tan^2 x + 1 = sec^2 x

    ---

    Number 8 also seems fine. (EDIT: Except you left out a negative sign on the RHS! )
    Last edited by janvdl; November 22nd 2007 at 10:04 AM. Reason: Oh that negative sign... :-(
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Member
    Joined
    Aug 2007
    Posts
    182
    What do I do to finish 7, bc the original equation was
    tan^2x + 5 = sec^2x + 4

    but I have only got:
    tan^2x + 1 = sec^2x
    Follow Math Help Forum on Facebook and Google+

  4. #4
    Bar0n janvdl's Avatar
    Joined
    Apr 2007
    From
    South Africa
    Posts
    1,630
    Thanks
    6
    Quote Originally Posted by aikenfan View Post
    What do I do to finish 7, bc the original equation was
    tan^2x + 5 = sec^2x + 4

    but I have only got:
    tan^2x + 1 = sec^2x
    But that is an identity...

    Oh i think i know what you mean:

    tan^2x + 1 = sec^2x

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

    [Multiply straight through with cos^2 x]
    Follow Math Help Forum on Facebook and Google+

  5. #5
    Bar0n janvdl's Avatar
    Joined
    Apr 2007
    From
    South Africa
    Posts
    1,630
    Thanks
    6
    For number 8:

    <br /> <br />
1 - \frac{1}{sin^2 x} = \frac{cos^2 x}{sin^2 x}<br />

    [Multiply straight through with sin^2 x here]

    Just check your Number 8, seems like you left out a negative sign on the right hand side.
    Follow Math Help Forum on Facebook and Google+

  6. #6
    Member
    Joined
    Aug 2007
    Posts
    182
    For number 6, I am supposed to turn the left side of the equation and turn it into the right side, using Quotient properties, etc...that is where I am stuck...I am not quite following...sorry for any confusion...I am having a really hard time with these...I don't think that I am done on 5 yet, because the left side is not equal to the right side.
    Follow Math Help Forum on Facebook and Google+

  7. #7
    Bar0n janvdl's Avatar
    Joined
    Apr 2007
    From
    South Africa
    Posts
    1,630
    Thanks
    6
    Quote Originally Posted by aikenfan View Post
    For number 6, I am supposed to turn the left side of the equation and turn it into the right side, using Quotient properties, etc...that is where I am stuck...I am not quite following...sorry for any confusion...I am having a really hard time with these...I don't think that I am done on 5 yet, because the left side is not equal to the right side.
    You're right, you are confusing me.

    sin^2 x + cos^2 x = 1

    For any value of x, you will get:

    1 = 1
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Simplifying a trigonometric expression.
    Posted in the Trigonometry Forum
    Replies: 1
    Last Post: October 16th 2010, 12:52 PM
  2. Replies: 6
    Last Post: August 29th 2010, 05:23 PM
  3. Simplifying Trigonometric Functions
    Posted in the Trigonometry Forum
    Replies: 7
    Last Post: March 8th 2010, 09:28 PM
  4. Simplifying Inverse Trigonometric Functions
    Posted in the Trigonometry Forum
    Replies: 3
    Last Post: July 10th 2009, 08:17 PM
  5. Simplifying trigonometric functions
    Posted in the Trigonometry Forum
    Replies: 12
    Last Post: November 18th 2007, 06:53 PM

Search Tags


/mathhelpforum @mathhelpforum