Results 1 to 5 of 5

Math Help - Trigonometry

  1. #1
    Junior Member
    Joined
    Aug 2008
    Posts
    65

    Trigonometry

    Greetings,

    I have the problem below. I've tried dividing everything to sina or cosa - I get one or several tga's, but eventually I end up with a single sina or cosa with some other numbers and from there I loop...

    Thank you.
    Attached Thumbnails Attached Thumbnails Trigonometry-trig.gif  
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Member
    Joined
    Aug 2008
    Posts
    80
    Hello,

    If tga means tan(a), use the formula 1+\tan^2 a=\frac{1}{\cos^2 a} and \cos^2 a+\sin^2 a=1.

    Bye
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Junior Member
    Joined
    Aug 2008
    Posts
    65
    I do not understand. I am left with an extra cosa again.

    Indeed, tg is tan. I am thought to abreviate it as tg.
    Attached Thumbnails Attached Thumbnails Trigonometry-trig.gif  
    Follow Math Help Forum on Facebook and Google+

  4. #4
    Member
    Joined
    Aug 2008
    Posts
    80
    Hello,

    Since tan a=-2 and 1+\tan^2 a=\frac{1}{\cos^2 a}, \cos a=\pm\frac{1}{\sqrt{5}}. Use \tan a=\frac{\sin a}{\cos a} to get \sin a=\mp\frac{1}{\sqrt{5}}. Plug in everything.

    Bye.
    Follow Math Help Forum on Facebook and Google+

  5. #5
    MHF Contributor
    Joined
    Apr 2005
    Posts
    1,631
    Quote Originally Posted by Logic View Post
    Greetings,

    I have the problem below. I've tried dividing everything to sina or cosa - I get one or several tga's, but eventually I end up with a single sina or cosa with some other numbers and from there I loop...

    Thank you.
    If the trigonometric functions don't want to cooperate, bypass them and use the reference triangle of the angle instead.

    tan(a) = -2
    That means "a" is in the 2nd or 4th quadrants.

    In the 2nd quadrant:
    opposite side = 2
    adjacent side = -1
    hypotenuse = sqrt(5)
    sin(a) = 2/sqrt(5)
    cos(a) = -1/sqrt(5)
    So,
    sin(a) / [sin^3(a) +3cos^2(a)
    = [2/sqrt(5)] / [(2/sqrt(5))^3 +3(-1/sqrt(5))^2]
    = 0.67989 -------------answer.

    In the 4th quadrant:
    opposite side = -2
    adjacent side = 1
    hypotenuse = sqrt(5)
    sin(a) = -2/sqrt(5)
    cos(a) = 1/sqrt(5)
    So,
    sin(a) / [sin^3(a) +3cos^2(a)
    = [-2/sqrt(5)] / [(-2/sqrt(5))^3 +3(1/sqrt(5))^2]
    = 7.74116 -------------answer also.
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Trigonometry to Memorize, and Trigonometry to Derive
    Posted in the Trigonometry Forum
    Replies: 9
    Last Post: August 21st 2013, 01:03 PM
  2. How To Do Trigonometry For This...?
    Posted in the Math Topics Forum
    Replies: 1
    Last Post: January 10th 2009, 06:56 PM
  3. How To Do This Trigonometry
    Posted in the Trigonometry Forum
    Replies: 2
    Last Post: January 3rd 2009, 02:55 AM
  4. trigonometry
    Posted in the Trigonometry Forum
    Replies: 3
    Last Post: December 31st 2008, 09:06 PM
  5. Trigonometry
    Posted in the Trigonometry Forum
    Replies: 2
    Last Post: December 18th 2008, 05:40 PM

Search Tags


/mathhelpforum @mathhelpforum