Question

Y= cos3x + 2sin3x. Express the function as a sinusoid in the form y= asinx (bx+c)

Answer 1

$a\sin(bx+c)=a\sin(bx)\cos(c)+a\cos(bx)\sin(c)=\cos(3x)+2\sin(3x)$

Clearly, you have $b=3$. In the second term of the middle expression, you need to have $a\sin(c)=1$, and in the first term, $a\cos(c)=2$.

$a\sin(c)=1\implies a=\dfrac1{\sin(c)}\implies a\cos(c)=\dfrac{\cos(c)}{\sin(c)}=\cot(c)=2$

which means $c=\mathrm{arccot}(2)$. Then,

$a\sin(\mathrm{arccot}(2))=\dfrac a{\sqrt5}=1\implies a=\sqrt5$

So,

$y=\cos(3x)+2\sin(3x)=\sqrt5\sin(3x+\mathrm{arccot}(2))$