Results 1 to 3 of 3

Math Help - Trigonometry

  1. #1
    Member
    Joined
    Sep 2009
    Posts
    98

    Trigonometry

    a) Sketch the graphs of y= tanA and y= secA

    b) Prove that secA- tanA= 1 / (secA+tanA)

    c) Deduce from parts a) and b) that secA-tanA is greater than 0 and less and equal to 1,

    for values of A such that A is greater and equal to 0 and less than and equal to 90 degrees. Explain your reasoning clearly.

    d) Given that secA-tanA= 1/2

    find the exact value of cosA
    Follow Math Help Forum on Facebook and Google+

  2. #2
    MHF Contributor
    Joined
    Sep 2008
    From
    West Malaysia
    Posts
    1,261
    Quote Originally Posted by Oasis1993 View Post
    a) Sketch the graphs of y= tanA and y= secA

    b) Prove that secA- tanA= 1 / (secA+tanA)

    c) Deduce from parts a) and b) that secA-tanA is greater than 0 and less and equal to 1,

    for values of A such that A is greater and equal to 0 and less than and equal to 90 degrees. Explain your reasoning clearly.

    d) Given that secA-tanA= 1/2

    find the exact value of cosA
    HI

    (b) \frac{1}{\sec A+\tan A}

    \frac{\cos A}{1+\sin A}

    \frac{\cos A}{1+\sin A}\cdot \frac{1-\sin A}{1-\sin A}

    \frac{\cos A(1-\sin A)}{1-\sin^2 A}

    \frac{\cos A(1-\sin A)}{\cos^2 A}

    some cancellings .. \sec A -\tan A

    (d) \sec A-\tan A=\frac{1}{2}

    \frac{1-\sin A}{\cos A}=\frac{1}{2}

    \cos A+2\sin A=2

    r\cos (A-\alpha)=2 *

    where r=\sqrt{1^2+2^2} and

    \tan \alpha=2\Rightarrow \alpha=\arctan 2

    evaluate r and alpha respectively and put it into the equation * .
    Follow Math Help Forum on Facebook and Google+

  3. #3
    MHF Contributor
    Joined
    Sep 2008
    From
    West Malaysia
    Posts
    1,261
    Quote Originally Posted by Oasis1993 View Post
    a) Sketch the graphs of y= tanA and y= secA

    b) Prove that secA- tanA= 1 / (secA+tanA)

    c) Deduce from parts a) and b) that secA-tanA is greater than 0 and less and equal to 1,

    for values of A such that A is greater and equal to 0 and less than and equal to 90 degrees. Explain your reasoning clearly.

    d) Given that secA-tanA= 1/2

    find the exact value of cosA
    HI

    from (b) , secA- tanA= 1 / (secA+tanA)

    \frac{1}{\sec A+\tan A}

    \frac{\cos A}{1+\sin A}

    When \frac{\cos A}{1+\sin A}

    is max , its denominator would be minimum ie sin A=0 , correspondingly cos A=1


    \frac{1}{1+0}=1

    Similarly , when its min , denominator would be max , ie sin A=1, cos A=0 ,

    \frac{0}{2}=0

    so we can see that 0\leq \sec A-\tan A\leq 1
    Attached Thumbnails Attached Thumbnails Trigonometry-trigo-graphs.jpg  
    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