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=\frac1{\sin(c)}\implies a\cos(c)=(\cos(c))/(\sin(c))=\cot(c)=2`

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


`a\sin(\mathrm{arccot}(2))=\frac a{\sqrt5}=1\implies a=\sqrt5`

So,


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