# Math Help - sin A+ sin B=x

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

2. Hello, tracker10!

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

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

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

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]

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

$\text{Add:}$

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

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?

3. Originally Posted by tracker10
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

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

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

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

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

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

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

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

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

4. $\displaystyle x=\sin(A)+\sin(B)=2\sin\left(\frac{A+B}{2}\right)\ cos\left(\frac{A-B}{2}\right),$ 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.

5. 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