Results 1 to 5 of 5

Math Help - sin A+ sin B=x

  1. #1
    Newbie
    Joined
    Feb 2011
    Posts
    1

    sin A+ sin B=x

    hello there iwas wandering if someone out there could help me.

    i have been given a problem and i am really struggling to resolve a solution

    if sin A + sin B = X
    cos A + cos B = y

    then prove sin (A+B)= (2xy)/(x^2+y^2)

    if you can help thankyou

    i have been trying to use the formula
    a sin x+ b cos x = y =cos (x+z) to no avail haha
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Super Member

    Joined
    May 2006
    From
    Lexington, MA (USA)
    Posts
    11,689
    Thanks
    617
    Hello, tracker10!

    I was off to a great start . . . then hit a wall.


    \text{if }\:\begin{Bmatrix}\sin A + \sin B \:=\: x & [1] \\ <br />
\cos A + \cos B \:=\: y & [2] \end{Bmatrix}

    . . \text{then prove: }\:\sin(A+B)\:=\: \dfrac{2xy}{x^2+y^2}

    Multiply [1] and [2]:

    . . \sin A\cos A + \sin A\cos B + \cos A\sin B + \sin B\cos B \:=\:xy


    We have:

    . . \underbrace{(\sin A\cos B + \cos A\sin B)}_{\text{This is }\sin(A+B)} + (\sin A\cos A + \sin B\cos B) \;=\;xy

    . . . . . . . . . . . . \sin(A+B) + \underbrace{(\sin A\cos A + \sin B\cos B)}_{\text{Now we want this.}} \;=\;xy .[1]


    \begin{array}{ccccccc}<br />
\text{Square [1]:} & \sin^2\!A + 2\sin A\sin B + \sin^2\!B &=& x^2 \\<br />
\text{Square [2]:} & \cos^2\!A + 2\cos A\cos B + \cos^2\!B &=& y^2 \end{array}


    \text{Add:}

    \underbrace{(\sin^2\!A+\cos^2\!A)}_{\text{This is 1}} + 2(\sin A\sin B + \cos A\cos B) + \underbrace{(\sin^2\!B + \cos^2\!B)}_{\text{This is 1}} \;=\;x^2+y^2

    And we have: . 2(\sin A\sin B + \cos A\cos B) + 2 \;=\;x^2y^2

    . . . \sin A\sin B + \cos A\cos B \;=\;\dfrac{x^2+y^2-2}{2}


    I intended to substitute this into [1]

    . . but it does not produced the desired result.


    Now what?

    Follow Math Help Forum on Facebook and Google+

  3. #3
    MHF Contributor
    Joined
    Dec 2009
    Posts
    3,120
    Thanks
    1
    Quote Originally Posted by tracker10 View Post
    hello there iwas wandering if someone out there could help me.

    i have been given a problem and i am really struggling to resolve a solution

    if sin A + sin B = X
    cos A + cos B = y

    then prove sin (A+B)= (2xy)/(x^2+y^2)

    if you can help thankyou

    i have been trying to use the formula
    a sin x+ b cos x = y =cos (x+z) to no avail haha
    SinA+SinB=x;\;\;\;\;\;CosA+CosB=y

    Prove that ....\;\;\;\;\;\;\;\displaystyle\ Sin(A+B)=\frac{2xy}{x^2+y^2}


    Sin(A+B)=SinACosB+CosASinB

    All you have to do is to calculate the numerator of \displaystyle\ ....\;\;\;\;\;\;\;Sin(A+B)=Sin(A+B)\left(\frac{x^2  +y^2}{x^2+y^2}\right)

    and show that it is 2xy

    \Rightarrow\left(SinACosB+CosASinB\right)\left[\frac{(SinA+SinB)^2+(CosA+CosB)^2}{(SinA+SinB)^2+(  CosA+CosB)^2}\right]

    =\frac{\left(SinACosB+CosASinB\right)\left[Sin^2A+Sin^2B+2SinASinB+Cos^2A+Cos^2B+2CosACosB\ri  ght]}{x^2+y^2}

    =\displaystyle\frac{\left(SinACosB+CosASinB\right)  \left[2+2SinASinB+2CosACosB\right]}{x^2+y^2}

    =\frac{2\left[SinACosB+CosASinB+Sin^2ASinBCosB+Sin^2BSinACosA+Si  nACosACos^2B+SinBCosBCos^2A\right]}{x^2+y^2}

    =\frac{2\left[SinACosB+CosASinB+\left(Sin^2A+Cos^2A\right)SinBCo  sB+\left(Sin^2B+Cos^2B\right)SinACosA\right]}{x^2+y^2}

    =\displaystyle\frac{2\left[SinACosB+CosASinB+SinBCosB+SinACosA\right]}{x^2+y^2}

    =\displaystyle\frac{2\left[SinA(CosA+CosB)+SinB(CosA+CosB)\right]}{x^2+y^2}

    =\displaystyle\frac{2\left[(SinA+SinB)(CosA+CosB)\right]}{x^2+y^2}=\frac{2xy}{x^2+y^2}
    Last edited by Archie Meade; February 3rd 2011 at 03:39 PM. Reason: explain strategy
    Follow Math Help Forum on Facebook and Google+

  4. #4
    A Plied Mathematician
    Joined
    Jun 2010
    From
    CT, USA
    Posts
    6,318
    Thanks
    4
    Awards
    2
    \displaystyle x=\sin(A)+\sin(B)=2\sin\left(\frac{A+B}{2}\right)\  cos\left(\frac{A-B}{2}\right), and similarly

    \displaystyle y=\cos(A)+\cos(B)=2\cos\left(\frac{A+B}{2}\right)\  cos\left(\frac{A-B}{2}\right).

    Therefore, we have that

    \displaystyle\frac{2xy}{x^{2}+y^{2}}=\frac{8\sin\l  eft(\frac{A+B}{2}\right)\cos\left(\frac{A+B}{2}\ri  ght)\cos^{2}\left(\frac{A-B}{2}\right)}{4\sin^{2}\left(\frac{A+B}{2}\right)\  cos^{2}\left(\frac{A-B}{2}\right)+4\cos^{2}\left(\frac{A+B}{2}\right)\c  os^{2}\left(\frac{A-B}{2}\right)}

    \displaystyle=\frac{8\sin\left(\frac{A+B}{2}\right  )\cos\left(\frac{A+B}{2}\right)\cos^{2}\left(\frac  {A-B}{2}\right)}{4\cos^{2}\left(\frac{A-B}{2}\right)}

    \displaystyle=2\sin\left(\frac{A+B}{2}\right)\cos\  left(\frac{A+B}{2}\right)

    \displaystyle=\sin(A+B).

    QED.
    Follow Math Help Forum on Facebook and Google+

  5. #5
    Banned
    Joined
    Dec 2010
    Posts
    6
    Or you can start it from right hand side. put x= (sinA + sinB) and y= (cosA + cosB). and solve it according to the equation and at last you will get LHS, hence proved. Thanks
    Follow Math Help Forum on Facebook and Google+

Search Tags


/mathhelpforum @mathhelpforum