1. ## Sin and Cos

How do I solve this? I have been working on this same problem for about an hour and can't come up with anything. I have to find the value of A, w, and b.

2sin(12x + 1) + 3sin(12x + 3) = A sin(wx + b)

Any help (with steps included) would be GREATLY appreciated! Thanks!

2. ok ... found an identity at this link that may help.

List of trigonometric identities - Wikipedia, the free encyclopedia

$a\sin{x} + b\sin(x+\alpha) = c\sin(x+\beta)$

where ...

$c = \sqrt{a^2+b^2+2ab\cos{\alpha}}$

and

$\beta = \arctan\left(\frac{b\sin{\alpha}}{a+b\cos{\alpha}} \right) + ...$

$0$ if $a+b\cos{\alpha} \ge 0$

$\pi$ if $a+b\cos{\alpha} < 0$

I'd let $x = 12\theta + 1$ and see where that goes.

3. Hello, aeagan2!

Wow . . . This took a lot of work!

I assume you know the Compound Angle Identities:

. . $\sin(A\pm B) \;=\;\sin A\cos B \pm \cos A\sin B$

. . $\cos(A \pm B) \;=\;\cos A\cos B \mp \sin A\sin B$

$2\sin(12x +1) + 3\sin(12x+3) = A\sin(wx+b)$

$\text{Find }A,\,w,\text{ and }b.$

We have:

. . $2\bigg[\sin12x\cos1 + \cos12x\sin1\bigg] + 3\bigg[\sin12x\cos3 + \cos12x\sin3\bigg]$
. . . . . . . . . . $=\;A\bigg[\sin wx\cos b + \cos wx\sin b\bigg]$

. . $2\cos1\sin12x + 2\sin1\cos12x + 3\cos3\sin12x + 3\sin3\cos12x$

. . . . . . . . . . $=\;A\cos b\sin wx + A\sin b\cos wx$

. . $(2\cos1 + 3\cos3)\sin12x + (2\sin1 + 3\sin3)\cos12x$

. . . . . . . . . . $=\;A\cos b\sin wx + A\sin b\cos 2x$

Equate "corresponding parts":

. . $\begin{array}{cccc}(2\cos1 + 3\cos3)\sin12x &=& A\cos b\sin wx & [1] \\ (2\sin 1 + 3\sin 3)\cos12x &=& A\sin b\cos wx & [2]\end{array}$

Let: . $\begin{Bmatrix}\sin12x &=& \sin wx \\ \cos12x &=& \cos wx\end{Bmatrix} \quad\Rightarrow\quad \boxed{w \:=\:12}$

[2] and [1] becomes: . $\begin{array}{cccc}A\sin b &=&2\sin1 + 3\sin3 & [3] \\A\cos b &=& 2\cos1 + 3\cos 3 & [4] \end{array}$

Divide [3] by [4]: . $\dfrac{A\sin b}{A\cos b} \;=\;\dfrac{2\sin1+3\sin3}{2\cos1+3\cos3} \;=\;\tan b$

. . $\text{Hence: }\;\boxed{b \;=\;\tan^{-1}\left(\frac{2\sin1+3\sin3}{2\cos1+3\cos3}\right) }$

$\begin{array}{cccccccc}\text{Square [3]:} & A^2\sin^2b &\!\!=\!\!& (2\sin 1 + 3\sin3)^2 &\!\!=\!\!& 4\sin^21 + 12\sin1\sin3 + 9\sin^23
\\ \text{Square [4]:} & A^2\cos^2b &\!\!=\!\!& (2\cos1+3\cos3)^2 &\!\!=\!\!& 4\cos^21 + 12\cos1\cos3 + 9\cos^23 \end{array}$

$A\underbrace{(\sin^2\!b+\cos^2\!b)}_{\text{This is 1}} \;=\;4\underbrace{(\sin^2\!1+\cos^2\!1)}_{\text{Th is is 1}} + 12\underbrace{(\cos3\cos1 + \sin3\sin1)}_{\text{This is }\cos(3-1)} + 9\underbrace{(\sin^2\!3+\cos^2\!3)}_{\text{This is 1}}$
And we have: . $A^2 \;=\;4 + 12\cos 2 + 9$
. . . . Hence: . $\boxed{A \;=\;\sqrt{13 + 12\cos2}}$