Question

Hi, I'm working through Problem Set 1 and I'm stuck on 1A-6. I've looked at the solutions and they don't make any sense. Can anyone explain the little trick they used to find the form A sin(x + c)?

Answer 1

Hi,
The "trick" is to expand the expression A sin(x+c) and match each term with the question. So, if we just get down to doing the same, we have
\[A \sin(x+c) = A (\sin(x)\cos(c) + \cos(x)\sin(c))\]

Matching each term with part a) of the question, it is easy to see that:
\[A \sin(x) \cos(c) = 1. \sin(x) . 1\] and \[A \cos(x) \sin(c) = 1. \cos(x) . \sqrt3\]

This gives, A cos(c) = 1, and A sin(c) = \[ \sqrt3\]

An easy way forward is to see that
\[\tan(c) = \frac{ A \sin(c) }{ A \cos(c) } = \sqrt{3}\]

Thus c = pi/3 (or 60 degrees).

Now, A is calculated as follows: A = 1 / cos(pi/3) = sqrt(3) / sin(pi/3) = 2! So, there you have the answer:
A sin (x + c) = 2 sin(x + pi/3)

The next part of the question should be solved similarly. Hope it helps!

Answer 2

Oops, I saw a typo, the second and third equation should be:
\[A \sin(x) \cos(c) = A \sin(x)\times1\] and \[A \cos(x) \sin(c) = A \cos(x)\times \sqrt{3}\]

Sorry about that!!