Question

Help with 1a 6b

Answer 1

I don't know if this is the fastest way, but here's what I would do:

\[A \sin(x+c) = A(sinx cosc + sinccosx)\]
Since there already is a sinx - cosx, we know that c will be in 4th quadrant because sinc must be negative and cosc must be positive, and that cosc = sinc (ignoring the signs)
This gives us \[3\pi/4\] or \[-\pi/4\]
And since we have a common factor of \[\sqrt{2}/2\] we need to make A to cancel that so A will = \[\sqrt{2}\]

And so we get:
\[\sqrt{2} \times (\sin(x+3\pi/4))\] or
\[\sqrt{2} \times (\sin(x-\pi/4))\]