Question

Hello, Can someone help me with the first problem set? In specific, I'm having problems with 1A6b and 1A7b. I think the answers here involve trig identities and graphing skills that I haven't used in a while. Can someone point me in the right direction? Thanks!

Answer 1

Can you post the specific question?

Answer 2

To express sin(X)-cos(x) in the form A sin(x+c):

You need to know the sum of angles formula for sin which was mentioned in lecture, with debate over the formula to emphasize that it has to be learned. It is sin(a+b) = sin(a)cos(b)+cos(a)sin(b)

Apply this to A sin (x+c)

Asin(x+c) = Asin(x)cos(c) + Acos(x)sin(c)

So if you want that to equal sin(x) - cos(x) what do the values of Acos(c) and Asin(c) have to be? You need the - sign on cos(x), so Asin(c) must be -1.

Now to find A you should have a value for Acos(c) and Asin(c).

Square them both and add them:

Asin(c) = -1
A^2sin^2(c) = (-1)(-1)
and square the value you get for Acos(c) the same way.

Now add these equations together:

A^2sin^2(c) + A^2cos^2(c) = whatever you got for the value of Asin(c) squared + 1 (which is the value Asin(c) squared.

This is A^2(sin^2(c) + cos^2(c)) = something

Now back to trig: one of the Pythagorean identities is:
sin^2(anything) + cos^2(that same thing) = 1.

So that makes A^2(1) = something

Now take the square root of both sides to find A

A = +/- sqrt(something)

Take just the positive root at first (you can verify later what happens if you take the negative root.

Now back there we had

Asin(c) = 1, but now we know A is the sqrt of something

sin(c) = 1/sqrt(something)

Likewise for what you found for cos(c)

now you know what the sine and cosine of c are.

You really should have memorized some of the values for key angles, but in this case you can draw a little triangle and make a good guess.

So you have

(whatyoufoundforA)sin(x+whatyoufoundforc) = sinx -cosx

As for the other one: you should be able to solve this by inspection. Yes, you should have memorized -cos(x+pi/2) = sin(x), but if you didn't doodle a little unit circle and figure it out. The the amplitude of sin(anything) is always 1 -- that is the absolute value of the most and least sin(x) can be.
So the amplitude of 4sin(x) is 4. And the rest of this is some basic trig stuff you should know but can figure out from your little unit circle. How far around the unit circle do you have to go to get all the possible values of sin(x) once? One time right? And how far is one time around a unit circle? 2pi, right? As for phase angle, how far out of phase with sin(x) is sin(x)? Zero, because it is not out of phase with itself.

Hope this helps.