Results 1 to 9 of 9

Thread: trigonometric identities

  1. #1
    Newbie
    Joined
    Apr 2008
    Posts
    12

    trigonometric identities

    I think I have this problem down but then again I'm not sure. If you could check my work, please:

    cot^2x = 1-2cscx+csc^2x
    csc^2x-1 = 1-2cscx+csc^2x (from the pythag. identity)
    0 = 2-2cscx + csc^2x-csc^2x
    0 = 2-2cscx
    0 = -1+cscx (divided all by -2)
    cscx = 1

    Also, I am so stuck on the following two. I have to use the double (or half, but I'm sure it's double) angle formulas but I don't know how to properly incorporate them:

    1. 2cos^2x-1 = cosx

    and

    2. cos2x + 6sin^2x = 2

    Help would be very much appreciated. Thank you!
    Last edited by augustfai; Apr 30th 2008 at 01:47 PM.
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Behold, the power of SARDINES!
    TheEmptySet's Avatar
    Joined
    Feb 2008
    From
    Yuma, AZ, USA
    Posts
    3,764
    Thanks
    78
    Quote Originally Posted by augustfai View Post
    Also, I am so stuck on the following two. I have to use the double angle formulas but I don't know how to properly incorporate them:

    1. 2cos^2x-1 = cosx

    and

    2. cos2x + 6sin^2x = 2

    Help would be very much appreciated. Thank you!
    For the first one this may help. let $\displaystyle u=\cos(x)$

    $\displaystyle \underbrace{2\cos^2(x)}_{2u^2}-1=\underbrace{\cos(x)}_{u}$

    $\displaystyle 2u^2-1=u \iff 2u^2-u-1=0 \iff (2u+1)(u-1)$

    so $\displaystyle u=-\frac{1}{2}$ and $\displaystyle u=1$

    but $\displaystyle u=cos(x)$ so we get

    $\displaystyle \cos(x)=-\frac{1}{2}$ and $\displaystyle \cos(x)=1$

    All that is left is to solve these.

    for the 2nd one use this identity
    using the identity $\displaystyle \cos(2x)=1-2\sin^{2}(x)$

    and try the sub $\displaystyle u=\sin(x)$

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

  3. #3
    MHF Contributor
    Joined
    Apr 2008
    Posts
    1,092
    Quote Originally Posted by augustfai View Post
    I think I have this problem down but then again I'm not sure. If you could check my work, please:

    cot^2x = 1-2cscx+csc^2x
    csc^2x-1 = 1-2cscx+csc^2x (from the pythag. identity)
    0 = 2-2cscx + csc^2x-csc^2x
    0 = 2-2cscx
    0 = -1+cscx (divided all by -2)
    cscx = 1

    Also, I am so stuck on the following two. I have to use the double angle formulas but I don't know how to properly incorporate them:

    1. 2cos^2x-1 = cosx

    and

    2. cos2x + 6sin^2x = 2

    Help would be very much appreciated. Thank you!
    Your work on the problem above is correct.

    For 1. $\displaystyle 2 \cos{^2}{x} - 1 = \cos{(2x)}$. I would show this using $\displaystyle \cos{(2x)} = \cos{(x+x)} = \cos{x}\cdot{\cos{x}} - \sin{x}\cdot{\sin{x}} = \cos{^2}{x} - \sin{^2}{x}$.
    Follow Math Help Forum on Facebook and Google+

  4. #4
    Newbie
    Joined
    Apr 2008
    Posts
    12
    Quote Originally Posted by icemanfan View Post
    Your work on the problem above is correct.

    For 1. $\displaystyle 2 \cos{^2}{x} - 1 = \cos{(2x)}$. I would show this using $\displaystyle \cos{(2x)} = \cos{(x+x)} = \cos{x}\cdot{\cos{x}} - \sin{x}\cdot{\sin{x}} = \cos{^2}{x} - \sin{^2}{x}$.
    Thank you, but I'm a little confused on this bit: how did $\displaystyle 2 \cos{^2}{x} - 1 = \cos{(x)}$ become $\displaystyle 2 \cos{^2}{x} - 1 = \cos{(2x)}$?
    Follow Math Help Forum on Facebook and Google+

  5. #5
    MHF Contributor
    Joined
    Apr 2008
    Posts
    1,092
    If the equation in 1 is written correctly, you may disregard my first post.

    For 2) use the double-angle identity that I worked out in the first post and substitute $\displaystyle 2 \cos{^2}{x} + 6 \sin{^2}{x} = 3$. Then:

    $\displaystyle 2 \cos{^2}{x} + 2 \sin{^2}{x} + 4 \sin{^2}{x} = 3$

    $\displaystyle 2 + 4 \sin{^2}{x} = 3$

    $\displaystyle 4 \sin{^2}{x} = 1$

    $\displaystyle \sin{^2}{x} = \frac{1}{4}$ and so

    $\displaystyle \sin{x} = \frac{1}{2}$ or $\displaystyle \sin{x} = -\frac{1}{2}$
    Follow Math Help Forum on Facebook and Google+

  6. #6
    MHF Contributor
    Joined
    Apr 2008
    Posts
    1,092
    Quote Originally Posted by augustfai View Post
    Thank you, but I'm a little confused on this bit: how did $\displaystyle 2 \cos{^2}{x} - 1 = \cos{(x)}$ become $\displaystyle 2 \cos{^2}{x} - 1 = \cos{(2x)}$?
    The equation that I worked out is an identity. It wasn't actually solving an equation. Since it looked so close to the identity I just assumed you had copied it incorrectly, but that was my mistake.
    Follow Math Help Forum on Facebook and Google+

  7. #7
    Newbie
    Joined
    Apr 2008
    Posts
    12
    Quote Originally Posted by icemanfan View Post
    The equation that I worked out is an identity. It wasn't actually solving an equation. Since it looked so close to the identity I just assumed you had copied it incorrectly, but that was my mistake.
    Ah, alright. Thank you. I actually don't know whether I'm supposed to use the double or the half-angle formulas, I assumed double since half doesn't seem to work out.

    EDIT:
    I think I sort of figured out the first one--

    $\displaystyle 2 \cos{^2}{x} - 1 = \cos{(x)}$

    turns into cos2x = cos x

    What, then, from there?
    Last edited by augustfai; Apr 30th 2008 at 02:03 PM.
    Follow Math Help Forum on Facebook and Google+

  8. #8
    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
    Hello,

    Quote Originally Posted by augustfai View Post
    Ah, alright. Thank you. I actually don't know whether I'm supposed to use the double or the half-angle formulas, I assumed double since half doesn't seem to work out.

    EDIT:
    I think I sort of figured out the first one--

    $\displaystyle 2 \cos{^2}{x} - 1 = \cos{(x)}$

    turns into cos2x = cos x

    What, then, from there?
    cos(a)=cos(b) <=> a=b+2k pi, OR a=-b+2k pi

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

  9. #9
    Newbie
    Joined
    Apr 2008
    Posts
    12
    I figured it out! Thanks, everyone.
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Trigonometric Identities
    Posted in the Trigonometry Forum
    Replies: 7
    Last Post: Jun 4th 2011, 10:08 PM
  2. Trigonometric identities
    Posted in the Pre-Calculus Forum
    Replies: 4
    Last Post: Dec 13th 2009, 06:48 PM
  3. Trigonometric Identities
    Posted in the Trigonometry Forum
    Replies: 2
    Last Post: Oct 6th 2009, 11:02 AM
  4. trigonometric identities
    Posted in the Trigonometry Forum
    Replies: 1
    Last Post: Nov 26th 2008, 03:03 PM
  5. trigonometric identities
    Posted in the Trigonometry Forum
    Replies: 3
    Last Post: May 8th 2007, 04:41 AM

Search Tags


/mathhelpforum @mathhelpforum