# Thread: Compound angles question Rsin...

1. ## Compound angles question Rsin...

Hello

I have a compound angles question that I have attempted but I am not sure if it is correct or not.

The question and my educated answer is laid out below:

Express 3sin(A) + 2cos(A) in the form, Rsin(A+α)

Now I have done the following after looking through books and teaching myself so it may not be right!

3sin(A) + 2cos(A) =Rsin(A+α)

=
R(sinA cosα + cosA sinα)

= (Rcosα) sinA + (Rsinα) cosA

Equating coefficients of sinA gives:

3 = Rcos
α , from which cosα = 3/R

Equating coefficients of cosA gives:

2= Rsinα , from which sinα = 2/R

R= (square root of) (3squared + 2 squared)

=
(square root of) 13

= 3.6

So so sorry about the confusion with the symbols and what not...

can anyone tell me if its right or if its not can they guide me in the right direction, sorry about the long winded post

Rob

2. Hello, Rob!

Your work looks good . . .

I have a compound angles question that I have attempted.

The question and my educated answer is laid out below:

Express $\displaystyle 3\sin A + 2\cos A$ in the form: $\displaystyle R\sin(A+\alpha)$

Now I have done the following after looking through books
and teaching myself so it may not be right!

$\displaystyle 3\sin A + 2\cos A \:=\:R\sin(A+\alpha)$

. . $\displaystyle = \;R(\sin A\cos\alpha + \cos A\sin\alpha)$

. . $\displaystyle = \;R\cos\alpha\sin A + R\sin\alpha \cos A$

Equating coefficients of $\displaystyle \sin A$ gives: .$\displaystyle R\cos\alpha \:=\:3 \quad\Rightarrow\quad \cos\alpha \:=\:\frac{3}{R}$

Equating coefficients of $\displaystyle \cos A$ gives: .$\displaystyle R\sin\alpha \:=\:2 \quad\Rightarrow\quad \sin\alpha \:=\:\frac{2}{R}$

. . Note that: .$\displaystyle {\color{blue}\tan\alpha \:=\:\frac{\sin\alpha}{\cos\alpha} \:=\:\frac{\frac{2}{R}}{\frac{3}{R}} \:=\:\frac{2}{3} \quad\Rightarrow\quad \alpha \:=\:\arctan(\tfrac{2}{3})}$

$\displaystyle R\:=\:\sqrt{3^2 + 2^2} \:=\:\sqrt{13} \:\approx\:3.6$

Can anyone tell me if its right? . . Ya done good!
. . $\displaystyle 3\sin A + 2\cos A \:=\:\sqrt{13}\sin(A + \alpha)\;\;\text{ where }\alpha = \arctan(\tfrac{2}{3})$