Page 1 of 2 12 LastLast
Results 1 to 15 of 18

Math Help - Trigonomic Equations

  1. #1
    Newbie UnstoppableBeast's Avatar
    Joined
    Aug 2011
    Posts
    12

    Trigonomic Equations

    cos^2(x)-cos2(x)+sin(x)=-sin^2(x)


    There's the problem, I've tried various different solutions but I can't seem to get it right, I've noticed that the middle cosine is a double-angle but I can't seem to make the formula give me an actual answer.

    For example, after canceling out and using the double-angle formula of cos2(x)=cos^2(x)-sin^2(x) I reached: -sin^2(x)+sin(x)=-sin^2(x) So...can that even be solved like that or have I done something wrong?


    csc^2(x)-csc(x)=2


    Same as the one above, despite the fact that it looks simpler, I'm basically incompetent. I have a slew of other problems similar to these and I think some examples will really help me figure out the others. Maybe there's some basic formula/identity that I'm missing out on.

    Note: I'm sorry for the title, I couldn't think of anything more specific. At least I'm following all the rules this time, right?
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Master Of Puppets
    pickslides's Avatar
    Joined
    Sep 2008
    From
    Melbourne
    Posts
    5,236
    Thanks
    28

    Re: Trigonomic Equations

    \cos^2(x)-\cos(2x)+\sin(x)=-\sin^2(x)

    \cos^2(x)+\sin^2(x)-\cos(2x)+\sin(x)=0

    \cos^2(x)+\sin^2(x)-(\cos^2(x)-\sin^2(x))+\sin(x)=0

    What happens next?
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Super Member Quacky's Avatar
    Joined
    Nov 2009
    From
    Windsor, South-East England
    Posts
    901

    Re: Trigonomic Equations

    You're on the right lines but I think I noticed a sign error with the original post. Using the identitity you stated, we get:

    cos^2(x)-[cos^2(x)-sin^2(x)]+sin(x)=-sin^2(x)
    Notice the signs? Check them below:
    cos^2(x)-cos^2(x)+sin^2(x)+sin(x)=-sin^2(x)

    sin^2(x)+sin(x)=-sin^2(x)

    2sin^2(x)+sin(x)=0

    Let t=sin(x)

    2t^2+t=0

    How would you solve that?

    Edit: Whoops, beaten to it, sorry!
    Follow Math Help Forum on Facebook and Google+

  4. #4
    Newbie UnstoppableBeast's Avatar
    Joined
    Aug 2011
    Posts
    12

    Re: Trigonomic Equations

    Ah I see, the initial part of the equation becomes "1" due to the pythagoreon identities but what I don't understand is what happens after. How do I get an actual answer in radians/degrees?
    Follow Math Help Forum on Facebook and Google+

  5. #5
    Master Of Puppets
    pickslides's Avatar
    Joined
    Sep 2008
    From
    Melbourne
    Posts
    5,236
    Thanks
    28

    Re: Trigonomic Equations

    Quote Originally Posted by Quacky View Post

    2t^2+t=0

    How would you solve that?

    ...
    Follow Math Help Forum on Facebook and Google+

  6. #6
    Newbie UnstoppableBeast's Avatar
    Joined
    Aug 2011
    Posts
    12

    Re: Trigonomic Equations

    2t^2=-t


    2t=-1


    t=-1/2

    Thanks for the help, it was a brain fart I suppose.

    Edit- Changing the sin/cos/tan into variables like "t" or "x" helped a lot, it's that kind of thinking/problem solving skill that I should've thought of. Often simplifying things makes it easier for my mind (and I assume everyone else's) to comprehend.
    Follow Math Help Forum on Facebook and Google+

  7. #7
    Super Member Quacky's Avatar
    Joined
    Nov 2009
    From
    Windsor, South-East England
    Posts
    901

    Re: Trigonomic Equations

    Quote Originally Posted by UnstoppableBeast View Post
    2t^2=-t


    2t=-1

    t=-1/2

    Thanks for the help, it was a brain fart I suppose.
    Ahhh! You need to do a quick review on algebra. Never divide by a variable - you lose a solution.

    2t^2+t=0

    Factoring:

    t(2t+1)=0

    t=0 or t=-0.5

    But you're not done. You know what t is. So what is x?
    Follow Math Help Forum on Facebook and Google+

  8. #8
    Newbie UnstoppableBeast's Avatar
    Joined
    Aug 2011
    Posts
    12

    Re: Trigonomic Equations

    O_O Wtf is wrong with me, I'm supposed to be going into an AP Calculus class as a 4.0 Student this year and I've forgotten everything.

    Anyways, I can handle the rest from here.

    sin(x)=0

    sin(x)=-.5



    x= 0, 11pi/6
    Follow Math Help Forum on Facebook and Google+

  9. #9
    MHF Contributor Siron's Avatar
    Joined
    Jul 2011
    From
    Norway
    Posts
    1,250
    Thanks
    20

    Re: Trigonomic Equations

    Don't forget there are infinite solutions.
    Follow Math Help Forum on Facebook and Google+

  10. #10
    Newbie UnstoppableBeast's Avatar
    Joined
    Aug 2011
    Posts
    12

    Re: Trigonomic Equations

    Thanks for your help everyone :3
    Follow Math Help Forum on Facebook and Google+

  11. #11
    MHF Contributor
    Prove It's Avatar
    Joined
    Aug 2008
    Posts
    11,548
    Thanks
    1418

    Re: Trigonomic Equations

    Quote Originally Posted by UnstoppableBeast View Post
    cos^2(x)-cos2(x)+sin(x)=-sin^2(x)


    There's the problem, I've tried various different solutions but I can't seem to get it right, I've noticed that the middle cosine is a double-angle but I can't seem to make the formula give me an actual answer.

    For example, after canceling out and using the double-angle formula of cos2(x)=cos^2(x)-sin^2(x) I reached: -sin^2(x)+sin(x)=-sin^2(x) So...can that even be solved like that or have I done something wrong?


    csc^2(x)-csc(x)=2


    Same as the one above, despite the fact that it looks simpler, I'm basically incompetent. I have a slew of other problems similar to these and I think some examples will really help me figure out the others. Maybe there's some basic formula/identity that I'm missing out on.

    Note: I'm sorry for the title, I couldn't think of anything more specific. At least I'm following all the rules this time, right?
    2.
    \displaystyle \begin{align*} \csc^2{x} - \csc{x} &= 2 \\ \csc^2{x} - \csc{x} - 2 &= 0 \\ X^2 - X - 2 &= 0 \textrm{ after making the substitution }X = \csc{x} \\ (X + 1)(X - 2) &= 0 \\ X + 1 = 0\textrm{ or }X - 2 &= 0 \\ X = -1\textrm{ or }X &=2 \end{align*}

    So solve \displaystyle \csc{x} = -1 and \displaystyle \csc{x} = 2
    Follow Math Help Forum on Facebook and Google+

  12. #12
    Member
    Joined
    Jul 2011
    Posts
    196

    Re: Trigonomic Equations

    Quote Originally Posted by Siron View Post
    Don't forget there are infinite solutions.
    So would it be:

    x = \pi k where k \in \mathbb{Z}

    x = \frac{7\pi}{6} + 2\pi k where k \in \mathbb{Z}

    x = \frac{11\pi}{6} + 2\pi k where k \in \mathbb{Z}
    Follow Math Help Forum on Facebook and Google+

  13. #13
    MHF Contributor Siron's Avatar
    Joined
    Jul 2011
    From
    Norway
    Posts
    1,250
    Thanks
    20

    Re: Trigonomic Equations

    @ terrorsquid: You forgot one solution!

    Solve \cos^2(x)-\cos(2x)+\sin(x)=-\sin^2(x)
    \Leftrightarrow \cos^2(x)-\cos(2x)+\sin(x)+\sin^2(x)=0
    \Leftightarrow 1-\cos(2x)+\sin(x)=0
    \Leftrightarrow 1-[1-2\sin^2(x)]+\sin(x)=0
    \Leftrightarrow 2\sin^2(x)+\sin(x)=0
    \Leftrightarrow \sin(x)[2\sin(x)+1]=0

    Two different cases:
    \sin(x)=0 \Leftrightarrow x=2k\pi \ \mbox{or} \ x=\pi + 2k\pi
    \sin(x)=\frac{-1}{2} \Leftrightarrow x=-\frac{\pi}{6}+2k\pi \ \mbox{or} \ x=\pi+\frac{\pi}{6}+2k\pi

    So four different solutions:
    V=\{2k\pi, \pi+2k\pi, \frac{-\pi}{6}+2k\pi, \frac{7\pi}{6}+2k\pi \}
    ( k\in \mathbb{Z})
    Follow Math Help Forum on Facebook and Google+

  14. #14
    Member
    Joined
    Jul 2011
    Posts
    196

    Re: Trigonomic Equations

    Isn't \{\pi k\} the same as \{2k\pi, \pi+2k\pi\} because starting from and including 0 + any integer multiple of \pi would =  \pi or 0?
    Follow Math Help Forum on Facebook and Google+

  15. #15
    MHF Contributor Siron's Avatar
    Joined
    Jul 2011
    From
    Norway
    Posts
    1,250
    Thanks
    20

    Re: Trigonomic Equations

    To check we write down a few solution of:
    \{k\pi \}=-\pi,0,\pi,2\pi,3\pi, ...
    \{2k\pi\}=-2\pi,0,2\pi,4\pi,...
    \{\pi +2k\pi\}=-\pi,\pi, 3\pi, 5\pi, ...

    So indeed that's an option, my apologies.
    Follow Math Help Forum on Facebook and Google+

Page 1 of 2 12 LastLast

Similar Math Help Forum Discussions

  1. [SOLVED] Help with trigonomic equation...
    Posted in the Trigonometry Forum
    Replies: 8
    Last Post: July 18th 2011, 07:23 PM
  2. Trigonomic graph
    Posted in the Trigonometry Forum
    Replies: 1
    Last Post: August 9th 2010, 11:16 AM
  3. Trigonomic Proof.
    Posted in the Trigonometry Forum
    Replies: 1
    Last Post: June 2nd 2010, 05:48 AM
  4. trigonomic identities
    Posted in the Trigonometry Forum
    Replies: 1
    Last Post: January 21st 2010, 02:04 PM
  5. Trigonomic Identities
    Posted in the Trigonometry Forum
    Replies: 1
    Last Post: June 20th 2007, 02:01 PM

Search Tags


/mathhelpforum @mathhelpforum