Results 1 to 7 of 7

Math Help - extremely frustrating identity...

  1. #1
    Member
    Joined
    Oct 2010
    From
    Heart of Winter
    Posts
    115
    Thanks
    1

    extremely frustrating identity...

    this one is giving me headache. I have tried maybe 7 tries and can not get close to what they want.

    express the fallowing in terms of a single function.

    sec \theta - tan \theta in terms of sin \theta .

    I can only use the elementary identities. Pythagorean Reciprocal Co-function even/odd.

    the answer to this in the book is  \frac{\sqrt{1-sin\theta}}{\sqrt{1+sin\theta}} which i can not get.

    2 questions...

    1) can I take the square root of the expression and multiply through the way you can with an equation? I know you can multiply by the conjugate of the denominator. but im not sure about the other two.

    2) I can't find the right strategy. I was thinking get a single term in numerator and denominator that equal part of the Pythagorean that have been squared. and go from there. but that seems to be long way around.

    any steps or ideas would be great.
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Super Member
    Joined
    Jun 2009
    Posts
    806
    Thanks
    4
    secθ - tanθ = 1/cosθ - sinθ/cosθ = (1 - sinθ)/cosθ
    Now substitute cosθ = sqrt( 1 - sin^2θ) and simplify.
    Follow Math Help Forum on Facebook and Google+

  3. #3
    A riddle wrapped in an enigma
    masters's Avatar
    Joined
    Jan 2008
    From
    Big Stone Gap, Virginia
    Posts
    2,551
    Thanks
    12
    Awards
    1
    Quote Originally Posted by skoker View Post
    this one is giving me headache. I have tried maybe 7 tries and can not get close to what they want.

    express the fallowing in terms of a single function.

    sec \theta - tan \theta in terms of sin \theta .

    I can only use the elementary identities. Pythagorean Reciprocal Co-function even/odd.

    the answer to this in the book is  \frac{\sqrt{1-sin\theta}}{\sqrt{1+sin\theta}} which i can not get.

    2 questions...

    1) can I take the square root of the expression and multiply through the way you can with an equation? I know you can multiply by the conjugate of the denominator. but im not sure about the other two.

    2) I can't find the right strategy. I was thinking get a single term in numerator and denominator that equal part of the Pythagorean that have been squared. and go from there. but that seems to be long way around.

    any steps or ideas would be great.
    Hi skoker,

    \sec \theta - \tan \theta = \frac{\sqrt{1-sin\theta}}{\sqrt{1+sin\theta}}

    Working from the left side...

    [1] \frac{1}{\cos \theta}-\frac{\sin \theta}{\cos \theta}=


    [2] \frac{1- \sin \theta}{\cos \theta} =

    Using \cos^2 \theta + \sin^2 \theta=1, we show that \cos \theta=\sqrt{1- \sin^2 \theta}

    [3] \frac{1- \sin \theta}{\sqrt{1- \sin^2 \theta}}=

    [4] \frac{1- \sin \theta}{\sqrt{1+\sin \theta} \cdot \sqrt{1-\sin \theta}}\cdot \frac{\sqrt{1-\sin \theta}}{\sqrt{1-\sin \theta}}=

    [5] \frac{(1-\sin \theta)\sqrt{1-\sin \theta}}{(1-\sin \theta)\sqrt{1+\sin \theta}}=

    [6]  \frac{\sqrt{1-sin\theta}}{\sqrt{1+sin\theta}}
    Last edited by masters; February 22nd 2011 at 09:45 AM. Reason: Added step numbers
    Follow Math Help Forum on Facebook and Google+

  4. #4
    Member
    Joined
    Oct 2010
    From
    Heart of Winter
    Posts
    115
    Thanks
    1
    AH! thank you masters and sa-ri-ga-ma.

    @masters
    so you split that denominator into

    \sqrt{1+sin\theta}\sqrt{1-sin\theta}

    because you factor the difference of 2 squares then split the factors?
    or because the root is congegate pair of -/+ ?

    that was the part i could not get. when i rationalized.
    Last edited by skoker; February 22nd 2011 at 10:04 AM.
    Follow Math Help Forum on Facebook and Google+

  5. #5
    Super Member

    Joined
    May 2006
    From
    Lexington, MA (USA)
    Posts
    11,751
    Thanks
    651

    I like sa-ri-ga-ma's suggestion!


    \displaystyle \sec\theta  - \tan\theta \;=\;\frac{1}{\cos\theta} - \frac{\sin\theta}{\cos\theta} \;=\; \frac{1 - \sin\theta}{\cos\theta}

    \text{Now substitute }\cos\theta \:=\: \sqrt{1 - \sin^2\!\theta}\,\text{ and simplify.}

    I assume "and simplify" means the following:

    \displaystyle \frac{1-\sin\theta}{\sqrt{1-\sin^2\!\theta}} \;=\;\frac{\left(\sqrt{1-\sin\theta}\right)^2} {\sqrt{(1-\sin\theta)(1+\sin\theta)}}

    . . . . . . . . . \displaystyle =\;\frac{\sqrt{1-\sin\theta}\,\sqrt{1-\sin\theta}}{\sqrt{1-\sin\theta}\,\sqrt{1+\sin\theta}}

    . . . . . . . . . \displaystyle =\;\frac{\sqrt{1-\sin\theta}}{\sqrt{1+\sin\theta}}

    Follow Math Help Forum on Facebook and Google+

  6. #6
    MHF Contributor

    Joined
    Apr 2005
    Posts
    15,792
    Thanks
    1532
    Quote Originally Posted by skoker View Post
    AH! thank you masters and sa-ri-ga-ma.

    @masters
    so you split that denominator into

    \sqrt{1+sin\theta}\sqrt{1-sin\theta}

    because you factor the difference of 2 squares then split the factors?
    Yes.

    or because the root is congegate pair of -/+ ?
    Isn't this really the same thing? What do you mean by "conjugate pair"?

    that was the part i could not get. when i rationalized.
    Follow Math Help Forum on Facebook and Google+

  7. #7
    Member
    Joined
    Oct 2010
    From
    Heart of Winter
    Posts
    115
    Thanks
    1
    Quote Originally Posted by HallsofIvy View Post

    What do you mean by "conjugate pair"?
    well this may be hard to explain because I'm not sure its the right terms or ideas but I will explain. I say conjugate pair meaning a pair of pos/neg values. like you have conjugate denominator and complex conjugate. and in the same way you have \pm a where a is a real number. which is the same as |a| which is I think same as \pm \sqrt{a}. but is not the same as \sqrt[3]{a} which is why you can have negative number under cube and only one principle root. little packages of +/- values. now I saw someone 'clear' out an absolute value with this (\pm a)^2=a this sort of struck me. and got me thinking about just what is going on with all this +/- business. and the relation to root and abs and +/-. so I see how this was split into +/- under a root and think for a second could you do that somehow other then factor? some technique I am missing or do not understand involving the fact that there is a square under a conjugate pair root? but then I see that it is a binomial which is more complicated to think about. the books do not go into this. they say you can clear a radical by raising to the power of the index. and this relates to the absolute value. but it seems that the relations between all these ideas are deeper.
    Last edited by skoker; February 23rd 2011 at 03:10 PM.
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Frustrating area calculation
    Posted in the Trigonometry Forum
    Replies: 1
    Last Post: May 23rd 2011, 02:27 AM
  2. Class boundaries-Frustrating
    Posted in the Statistics Forum
    Replies: 1
    Last Post: March 6th 2009, 08:06 AM
  3. Frustrating Hmwk DUE
    Posted in the Trigonometry Forum
    Replies: 2
    Last Post: October 3rd 2008, 04:10 PM
  4. Frustrating integral
    Posted in the Calculus Forum
    Replies: 3
    Last Post: December 3rd 2007, 10:56 PM
  5. Frustrating problem
    Posted in the Geometry Forum
    Replies: 3
    Last Post: June 29th 2007, 07:54 PM

Search Tags


/mathhelpforum @mathhelpforum