Results 1 to 3 of 3

Math Help - Exact value of Cos(Alpha-Beta)

  1. #1
    Junior Member
    Joined
    May 2009
    Posts
    39

    Exact value of Cos(Alpha-Beta)

    cosalpha = sqrt(3)/4 in quadrant I
    cosbeta = -(sqrt2)/3 in quadrant II

    Find the exact value of cos(alpha - beta)
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Master Of Puppets
    pickslides's Avatar
    Joined
    Sep 2008
    From
    Melbourne
    Posts
    5,236
    Thanks
    28
     cos(a-b) = cos(a)cos(b)+sin(a)sin(b)

    To find the values for sin(a) and sin(b) use sin^2(x)+cos^2(x)=1
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Super Member

    Joined
    May 2006
    From
    Lexington, MA (USA)
    Posts
    11,690
    Thanks
    617
    Hello, Neversh!

    Given: . \begin{array}{c}\cos\alpha \:=\: \dfrac{\sqrt{3}}{4}\;\text{ in Quadrant 1} \\ \\[-3mm]<br />
\cos\beta \:=\: \text{-}\dfrac{\sqrt{2}}{3}\;\text{ in Quadrant 2} \end{array}

    Find the exact value of: . \cos(\alpha - \beta)
    We know: . \sin^2\!\theta + \cos^2\!\theta \:=\:1 \quad\Rightarrow\quad \sin\theta \:=\:\pm\sqrt{1-\cos^2\!\theta}


    Since {\color{blue}\cos\alpha = \frac{\sqrt{3}}{4}}, then \sin\alpha \:=\:\pm\sqrt{1 - \left(\tfrac{\sqrt{3}}{4}\right)^2} \:=\:\pm\sqrt{1 - \tfrac{3}{16}} \:=\:\pm\sqrt{\tfrac{13}{16}} \:=\:\pm\tfrac{\sqrt{13}}{4}

    . . \alpha is in Quadrant 1, so: . {\color{blue}\sin\alpha \:=\:\frac{\sqrt{13}}{4}}


    Since {\color{blue}\cos\beta = \text{-}\frac{\sqrt{2}}{3}}, then \sin\beta\:=\:\pm\sqrt{1 - \left(\text{-}\tfrac{\sqrt{2}}{3}\right)^2} \:=\:\pm\sqrt{1-\tfrac{2}{9}} \:=\:\pm\sqrt{\tfrac{7}{9}} \:=\:\pm\tfrac{\sqrt{7}}{3}

    . . \beta is in Quadrant 2, so: . {\color{blue}\sin\beta \:=\:\frac{\sqrt{7}}{3}}


    We know: . \cos(\alpha - \beta ) \;=\;\cos\alpha\cos\beta + \sin\alpha\sin\beta


    Substitute the above four values:

    . . \cos(\alpha - \beta) \;=\;\left(\tfrac{\sqrt{3}}{4}\right)\left(\text{-}\tfrac{\sqrt{2}}{3}\right) + \left(\tfrac{\sqrt{13}}{4}\right)\left(\tfrac{\sqr  t{7}}{3}\right) \;=\;\text{-}\tfrac{\sqrt{6}}{12} + \tfrac{\sqrt{91}}{12}


    Therefore: . \cos(\alpha - \beta) \;=\;\frac{\sqrt{91} - \sqrt{6}}{12}

    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. |alpha-beta|
    Posted in the Algebra Forum
    Replies: 1
    Last Post: January 30th 2009, 06:28 AM
  2. derive cos(alpha+beta) sin(alpha+beta)?
    Posted in the Trigonometry Forum
    Replies: 3
    Last Post: December 7th 2008, 03:27 PM
  3. Solve for alpha and beta
    Posted in the Trigonometry Forum
    Replies: 2
    Last Post: September 18th 2007, 08:26 AM
  4. Values for alpha and beta.
    Posted in the Algebra Forum
    Replies: 6
    Last Post: July 2nd 2007, 03:50 PM
  5. FIND sin alpha and sin beta
    Posted in the Trigonometry Forum
    Replies: 3
    Last Post: April 17th 2006, 09:34 AM

Search Tags


/mathhelpforum @mathhelpforum