Results 1 to 7 of 7

Math Help - Trig

  1. #1
    Junior Member
    Joined
    Apr 2008
    Posts
    30

    Trig

    Heres the question, a full solution step by step is great as i already know the process to answer a question like this but have trouble doing so.\

     tanx-sinx/tanxsinx = tanxsinx/tanx+sinx
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Moo
    Moo is offline
    A Cute Angle Moo's Avatar
    Joined
    Mar 2008
    From
    P(I'm here)=1/3, P(I'm there)=t+1/3
    Posts
    5,618
    Thanks
    6
    Hi again !

    Quote Originally Posted by FORK View Post
    Heres the question, a full solution step by step is great as i already know the process to answer a question like this but have trouble doing so.\

     tanx-sinx/tanxsinx = tanxsinx/tanx+sinx
    I must guess it's \frac{\tan x-\sin x}{\tan x \sin x}=\frac{\tan x \sin x}{\tan x+\sin x}

    Please next time, put parenthesis or brackets...
    Now, do you have to prove this equation or solve for x ?
    And what about showing some of your steps ?
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Junior Member
    Joined
    Apr 2008
    Posts
    30
    Quote Originally Posted by Moo View Post
    Hi again !



    I must guess it's \frac{\tan x-\sin x}{\tan x \sin x}=\frac{\tan x \sin x}{\tan x+\sin x}

    Please next time, put parenthesis or brackets...
    Now, do you have to prove this equation or solve for x ?
    And what about showing some of your steps ?
    Sorry, i didnt know how to do that. And that equation must be proved. Thanks.

    Oh, and this is the equation i need to prove, i havent done any work to it yet. Help please.
    Last edited by FORK; June 16th 2008 at 02:51 PM.
    Follow Math Help Forum on Facebook and Google+

  4. #4
    Forum Admin topsquark's Avatar
    Joined
    Jan 2006
    From
    Wellsville, NY
    Posts
    9,854
    Thanks
    321
    Awards
    1
    Quote Originally Posted by FORK View Post
    Heres the question, a full solution step by step is great as i already know the process to answer a question like this but have trouble doing so.\

     tanx-sinx/tanxsinx = tanxsinx/tanx+sinx
    \frac{tan(x) - sin(x)}{tan(x)~sin(x)} = \frac{tan(x)~sin(x)}{tan(x) + sin(x)}
    One way is to start on the left hand side and convert everything to sines and cosines:
    = \frac{\frac{sin(x)}{cos(x)} - sin(x)}{\frac{sin(x)}{cos(x)}~sin(x)}

    Multiply top and bottom by cos(x):
    = \frac{\frac{sin(x)}{cos(x)} - sin(x)}{\frac{sin(x)}{cos(x)}~sin(x)} \cdot \frac{cos(x)}{cos(x)}

    = \frac{\frac{sin(x)}{cos(x)} \cdot cos(x) - sin(x)~cos(x)}{\frac{sin(x)}{cos(x)}~sin(x) \cdot cos(x)}

    = \frac{sin(x) - sin(x)~cos(x)}{sin^2(x)}

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

    = \frac{sin(x)(1 - cos(x))}{(1 - cos(x))(1 + cos(x))}

    = \frac{sin(x)}{1 + cos(x)}

    Now multiply the top and bottom by tan(x):
    = \frac{sin(x)}{1 + cos(x)} \cdot \frac{tan(x)}{tan(x)}

    = \frac{tan(x)~sin(x)}{tan(x) + tan(x)~cos(x)}

    = \frac{tan(x)~sin(x)}{tan(x) + sin(x)}

    -Dan
    Follow Math Help Forum on Facebook and Google+

  5. #5
    Junior Member
    Joined
    Apr 2008
    Posts
    30
    Quote Originally Posted by topsquark View Post
    \frac{tan(x) - sin(x)}{tan(x)~sin(x)} = \frac{tan(x)~sin(x)}{tan(x) + sin(x)}
    One way is to start on the left hand side and convert everything to sines and cosines:
    = \frac{\frac{sin(x)}{cos(x)} - sin(x)}{\frac{sin(x)}{cos(x)}~sin(x)}

    Multiply top and bottom by cos(x):
    = \frac{\frac{sin(x)}{cos(x)} - sin(x)}{\frac{sin(x)}{cos(x)}~sin(x)} \cdot \frac{cos(x)}{cos(x)}

    = \frac{\frac{sin(x)}{cos(x)} \cdot cos(x) - sin(x)~cos(x)}{\frac{sin(x)}{cos(x)}~sin(x) \cdot cos(x)}

    = \frac{sin(x) - sin(x)~cos(x)}{sin^2(x)}

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

    = \frac{sin(x)(1 - cos(x))}{(1 - cos(x))(1 + cos(x))}

    = \frac{sin(x)}{1 + cos(x)}

    Now multiply the top and bottom by tan(x):
    = \frac{sin(x)}{1 + cos(x)} \cdot \frac{tan(x)}{tan(x)}

    = \frac{tan(x)~sin(x)}{tan(x) + tan(x)~cos(x)}

    = \frac{tan(x)~sin(x)}{tan(x) + sin(x)}

    -Dan
    Wow man, thanks alot, I certainly appreciate it. Ill check this with my teacher tommorow, thanks alot again
    Follow Math Help Forum on Facebook and Google+

  6. #6
    Junior Member
    Joined
    Apr 2008
    Posts
    30
    Quote Originally Posted by topsquark View Post
    \frac{tan(x) - sin(x)}{tan(x)~sin(x)} = \frac{tan(x)~sin(x)}{tan(x) + sin(x)}
    One way is to start on the left hand side and convert everything to sines and cosines:
    = \frac{\frac{sin(x)}{cos(x)} - sin(x)}{\frac{sin(x)}{cos(x)}~sin(x)}

    Multiply top and bottom by cos(x):
    = \frac{\frac{sin(x)}{cos(x)} - sin(x)}{\frac{sin(x)}{cos(x)}~sin(x)} \cdot \frac{cos(x)}{cos(x)}

    = \frac{\frac{sin(x)}{cos(x)} \cdot cos(x) - sin(x)~cos(x)}{\frac{sin(x)}{cos(x)}~sin(x) \cdot cos(x)}

    = \frac{sin(x) - sin(x)~cos(x)}{sin^2(x)}

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

    = \frac{sin(x)(1 - cos(x))}{(1 - cos(x))(1 + cos(x))}

    = \frac{sin(x)}{1 + cos(x)}

    Now multiply the top and bottom by tan(x):
    = \frac{sin(x)}{1 + cos(x)} \cdot \frac{tan(x)}{tan(x)}

    = \frac{tan(x)~sin(x)}{tan(x) + tan(x)~cos(x)}

    = \frac{tan(x)~sin(x)}{tan(x) + sin(x)}

    -Dan
    COuple questions though, are you allowed to just multiply them by tan if youre not making the denominator common? And how did you get from your second last step to the last one?
    Follow Math Help Forum on Facebook and Google+

  7. #7
    Forum Admin topsquark's Avatar
    Joined
    Jan 2006
    From
    Wellsville, NY
    Posts
    9,854
    Thanks
    321
    Awards
    1
    Quote Originally Posted by FORK View Post
    COuple questions though, are you allowed to just multiply them by tan if youre not making the denominator common? And how did you get from your second last step to the last one?
    \frac{a}{b} \cdot \frac{c}{c} = \frac{ac}{bc} (with suitable restrictions on b and c.) When you multiply fractions you do not need common denominators.

    Finishing the problem is easy! C'mon, you need to try a little harder.

    \frac{tan(x)~sin(x)}{tan(x) + tan(x)~cos(x)}

    = \frac{tan(x)~sin(x)}{tan(x) + \frac{sin(x)}{cos(x)}~cos(x)}

    = \frac{tan(x)~sin(x)}{tan(x) + sin(x)}

    -Dan
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Compute Trig Function Values, Solve Trig Equation
    Posted in the Trigonometry Forum
    Replies: 1
    Last Post: September 8th 2011, 07:00 PM
  2. Replies: 7
    Last Post: April 15th 2010, 08:12 PM
  3. Replies: 6
    Last Post: November 20th 2009, 04:27 PM
  4. Replies: 1
    Last Post: July 24th 2009, 02:29 AM
  5. Trig Equations with Multiple Trig Functions cont.
    Posted in the Trigonometry Forum
    Replies: 4
    Last Post: April 7th 2008, 05:50 PM

Search Tags


/mathhelpforum @mathhelpforum