Results 1 to 5 of 5

Math Help - tan trig angle formula

  1. #1
    Senior Member
    Joined
    Nov 2008
    Posts
    425

    tan trig angle formula

    Determine an exact value for the expression:

    [sin(pi/3)-cos(5pi/6)]
    ----------------------
    [1-tan(3pi/4)cot(pi/4)]

    ok, here are my steps: can someone tell me where my errors are

    the answer is supposed to be sqrt3/2

    [sin(pi/3)-cos(5pi/6)]
    ----------------------
    [1-tan(3pi/4)cot(pi/4)]
    [(sqrt3/2)-(-sqrt3/2)]/[1-(-1)(1)
    [sqrt3/2+sqrt3/2]/2
    [sqrt9/2]/2
    sqrt3

    I seem to have lost the 2 in the denominator? Where is my error? thanks
    Last edited by skeske1234; March 29th 2009 at 12:27 PM.
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Member
    Joined
    Jan 2009
    Posts
    145
    Could you elaborate slightly on what your problem is? It looks rather straightforward. Do you have problems calculating which each of the trig functions are? If so, try drawing the unit circle.
    Follow Math Help Forum on Facebook and Google+

  3. #3
    MHF Contributor

    Joined
    Apr 2005
    Posts
    16,427
    Thanks
    1857
    Quote Originally Posted by skeske1234 View Post
    I don't understand the tan trig angle formula:

    So, how would I do this question?

    Determine an exact value for the expression:

    [sin(pi/3)-cos(5pi/6)]/[1-tan(3pi/4)cot(pi/4)]

    Can you give me all the details and steps please
    Just go ahead and calculate the individual trig values! For example, " \pi/3" is the same as 60 degrees so you can think of a 60-30 right triangle as half an equilateral triangle. If you take the hypotenuse to be "h" then the leg adjacent to the 60 degree angle has length h/2 and you can find the length of the other leg by using the Pythagorean theorem. Once you have the three lengths, you can get cos(\pi/3) and sin(\pi/3). For tan(4\pi/4) and cot(\pi/4), use the fact that a " \pi/4-\pi/4" or "45-45" triangle is isosceles. The "near side" and "opposite" side have the same length.
    Follow Math Help Forum on Facebook and Google+

  4. #4
    Senior Member
    Joined
    Nov 2008
    Posts
    425
    ok, here are my steps: can someone tell me where my errors are

    the answer is supposed to be sqrt3/2

    [sin(pi/3)-cos(5pi/6)]/[1-tan(3pi/4)cot(pi/4)]
    [(sqrt3/2)-(-sqrt3/2)]/[1-(-1)(1)
    [sqrt3/2+sqrt3/2]/2
    [sqrt9/2]/2
    sqrt3

    I seem to have lost the 2 in the denominator? Where is my error? thanks
    Follow Math Help Forum on Facebook and Google+

  5. #5
    Super Member

    Joined
    May 2006
    From
    Lexington, MA (USA)
    Posts
    11,909
    Thanks
    772
    Hello, skeske1234!

    Here are my steps:

    \frac{\sin\frac{\pi}{3} - \cos\frac{5\pi}{6}} {1-\tan\frac{3\pi}{4}\cot\frac{\pi}{4}}

    . .  =\; \frac{\dfrac{\sqrt{3}}{2}-\left(-\dfrac{\sqrt{3}}{2}\right)} {1-(-1)(1)}

    . . = \;\frac{\dfrac{\sqrt{3}}{2} +\dfrac{\sqrt{3}}{2}}{2}

    . . = \;\frac{\left(\dfrac{\sqrt{{\color{red}9}}}{2}\rig  ht)}{2} . . . nine?

    You know this, right? . \frac{\sqrt{3}}{2} + \frac{\sqrt{3}}{2} \;=\;\sqrt{3}

    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Double Angle Formula (Sin 2A, Cos 2A...)
    Posted in the Trigonometry Forum
    Replies: 3
    Last Post: September 18th 2011, 07:41 AM
  2. Trig Half-Angle Formula question
    Posted in the Trigonometry Forum
    Replies: 1
    Last Post: October 23rd 2010, 11:07 PM
  3. Replies: 7
    Last Post: April 15th 2010, 09:12 PM
  4. Finding angle/ Double angle formula
    Posted in the Trigonometry Forum
    Replies: 2
    Last Post: May 29th 2009, 08:43 AM
  5. Using half angle formula
    Posted in the Trigonometry Forum
    Replies: 6
    Last Post: March 27th 2006, 04:50 AM

Search Tags


/mathhelpforum @mathhelpforum