Results 1 to 11 of 11

Thread: Using identities

  1. #1
    Member Furyan's Avatar
    Joined
    Sep 2011
    Posts
    132
    Thanks
    3

    Using identities

    Hello,

    The question Iím working on is

    $\displaystyle f(x) = \4cos^2{x} - \3sin^2{x}$

    Show that $\displaystyle f(x) = \dfrac{1}{2} +\dfrac{7}{2}\cos2{x} $

    I have tried using $\displaystyle \sin^2{x} + \cos^2{x} = 1$

    And the double angle identity for $\displaystyle \cos2{x}$

    The best Iíve got is

    $\displaystyle f(x) = \3cos2{x} + \cos^2{x}$, which it might not be.

    Maybe I should be using the half angle identity, but I'm not sure how.

    I know I have to work at these, but if someone could point me in the right direction with this Iíd be very grateful.

    Thank you
    Last edited by Furyan; Dec 18th 2011 at 11:50 AM. Reason: Correction
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Super Member ILikeSerena's Avatar
    Joined
    Dec 2011
    Posts
    734
    Thanks
    122

    Re: Using identities

    Hi Furyan!

    Almost there!
    What is the double angle identity for $\displaystyle \cos 2x$ in terms of $\displaystyle \cos^2 x$?
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Member Furyan's Avatar
    Joined
    Sep 2011
    Posts
    132
    Thanks
    3

    Re: Using identities

    Hi ILikeSerena

    Almost there, really? Cool

    It's $\displaystyle \cos2{x} = \2cos^2{x} - 1$?

    I'll try using it and see what I get.

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

  4. #4
    MHF Contributor Siron's Avatar
    Joined
    Jul 2011
    From
    Belgium
    Posts
    1,254
    Thanks
    24

    Re: Using identities

    We have given:
    $\displaystyle f(x)=4\cos^2(x)-3\sin^2(x)$
    which we can write as:
    $\displaystyle f(x)=4\cos^2(x)-3[1-\cos^2(x)]$
    $\displaystyle =4\cos^2(x)-3+3\cos^2(x)$
    $\displaystyle =7\cos^2(x)-3$
    $\displaystyle =\frac{7}{2}\left[2\cos^2(x)-\frac{6}{7}]$
    $\displaystyle =\frac{7}{2}\left[2\cos^2(x)-1+\frac{1}{7}\right]$
    $\displaystyle =\frac{7}{2}\left[\cos(2x)+\frac{1}{7}\right]$
    $\displaystyle =\frac{7}{2}\cos(2x)+\frac{1}{2}$
    Last edited by Siron; Dec 18th 2011 at 12:37 PM.
    Follow Math Help Forum on Facebook and Google+

  5. #5
    Member Furyan's Avatar
    Joined
    Sep 2011
    Posts
    132
    Thanks
    3

    Re: Using identities

    Hi LikeSerena

    Thank you for letting me know I was on the right track, it really helped. I think I have it.

    $\displaystyle \cos2{x} = \2cos^2{x} - 1$

    $\displaystyle \cos2{x} + 1 = \2cos^2{x}$

    $\displaystyle \dfrac{1}{2}\(cos2{x}+1) =\cos^2{x} $

    Substituting I get

    $\displaystyle \3cos2{x}+ \dfrac{1}{2}(\cos2{x}+1) $

    $\displaystyle \3cos2{x}+ \dfrac{1}{2}\cos2{x}+ \dfrac{1}{2} $

    $\displaystyle \dfrac{7}{2}\cos2{x}+ \dfrac{1}{2} $

    That appears to be it. Yipee, thank you!
    Follow Math Help Forum on Facebook and Google+

  6. #6
    Super Member ILikeSerena's Avatar
    Joined
    Dec 2011
    Posts
    734
    Thanks
    122

    Re: Using identities

    Good!
    Follow Math Help Forum on Facebook and Google+

  7. #7
    Member Furyan's Avatar
    Joined
    Sep 2011
    Posts
    132
    Thanks
    3

    Re: Using identities

    Hi Siron

    $\displaystyle =7\cos^2(x)-3$

    This is another approach I had tried, but I had got completely stuck, my factoring is not so good. I am very grateful for your reply and I will enjoy working through your method. Thank you very much for taking the time to reply, I am very grateful. It is enormously helpful.
    Follow Math Help Forum on Facebook and Google+

  8. #8
    Member Furyan's Avatar
    Joined
    Sep 2011
    Posts
    132
    Thanks
    3

    Re: Using identities

    Last edited by Furyan; Dec 18th 2011 at 02:05 PM. Reason: duplicate
    Follow Math Help Forum on Facebook and Google+

  9. #9
    MHF Contributor
    Joined
    Dec 2009
    Posts
    3,120
    Thanks
    4

    Re: Using identities

    You could also try

    $\displaystyle cos2x=cos^2x-sin^2x$

    $\displaystyle 4cos^2x-3sin^2x=cos^2x+3\left(cos^2x-sin^2x\right)=cos^2x+3cos2x$

    $\displaystyle =\frac{2cos^2x}{2}-\frac{cos2x}{2}+\frac{7cos2x}{2}$

    $\displaystyle =\frac{2cos^2x-cos^2x+sin^2x}{2}+\frac{7cos2x}{2}=\frac{cos^2x+si n^2x}{2}+\frac{7cos2x}{2}$

    $\displaystyle =\frac{1}{2}+\frac{7}{2}cos2x$
    Follow Math Help Forum on Facebook and Google+

  10. #10
    Member Furyan's Avatar
    Joined
    Sep 2011
    Posts
    132
    Thanks
    3

    Re: Using identities

    Thank you ILikeSerena You are great teacher!
    Last edited by Furyan; Dec 18th 2011 at 02:24 PM. Reason: Correction
    Follow Math Help Forum on Facebook and Google+

  11. #11
    Member Furyan's Avatar
    Joined
    Sep 2011
    Posts
    132
    Thanks
    3

    Re: Using identities

    Thank you very much Archie Meade I will try that too.

    Awesome!
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Replies: 6
    Last Post: Jun 22nd 2010, 11:59 PM
  2. identities..
    Posted in the Trigonometry Forum
    Replies: 3
    Last Post: Jul 29th 2009, 12:33 PM
  3. Identities
    Posted in the Trigonometry Forum
    Replies: 2
    Last Post: Jul 14th 2009, 07:58 AM
  4. identities
    Posted in the Trigonometry Forum
    Replies: 5
    Last Post: Nov 2nd 2008, 03:34 PM
  5. Identities
    Posted in the Trigonometry Forum
    Replies: 3
    Last Post: Jul 11th 2008, 03:53 AM

Search Tags


/mathhelpforum @mathhelpforum