# Thread: Rewriting an equation into the form of Sin(a+b)

1. ## Rewriting an equation into the form of Sin(a+b)

How do I rewrite y = sinx + (sqrt(3))*(cosx) into the form of sinx(a+b)?

2. $y=sin x+\sqrt{3}cos x$
$y=2(\frac{1}{2}sin x+\frac{\sqrt 3}{2}cos x)$
$y=2(sin 60 cos x + cos 60 sin x)$
$y=2sin (x+60)$

3. In general, if you have Asin(x)+ Bcos(x) you can write it as either a sine or cosine function.

sin(a+b)= sin(a)cos(b)+ cos(a)sin(b) so if you can find "b" so that cos(b)= A and sin(b)= B you have it. Of course, in that case, $A^2+ B^2= cos^2(b)+ sin^2(b)= 1$ so you must first have $A^2+ B^2= 1$. In your problem, A= 1 and $B= \sqrt{3}$ so $A^2+ B^2= 1+ 3= 4$, not 1. That was why alexmahone factored out $2= \sqrt{4}$ first.

For any A sin(x)+ B cos(x), We can write $A sin(x)+ B cos(x)= \sqrt{A^2+ B^2}\left(\frac{A}{\sqrt{A^2+ B^2}}sin(x)+ \frac{B}{/sqrt{A^2+ B^2}}cos(x)\right)$ $= \sqrt{A^2+ B^2}sin(x+ \theta)$ where $cos(\theta)= \frac{A}{\sqrt{A^2+ B^2}}$ and $sin(\theta)= \frac{B}{\sqrt{A^2+ B^2}}$. Since $tan(\theta)= \frac{sin(\theta)}{cos(\theta)}= \frac{B}{A}$, $\theta= \arctan\left(\frac{B}{A}\right)$

To write A sin(x)+ B cos(x) as a cosine, use cos(a+b)= cos(a)cos(b)- sin(a)sin(b) and do the same thing.