1. We are given two points that are on the graph of \(y = -2 \sin(3x) + 1\), which are \((-\frac{\pi}{12}, a) \) and \((b, 1)\). Find a and b.

2. We are given two points that are on the graph of \(y = \tan(2x + \pi/2)\), which are \((-\frac{5\pi}{12}, a)\) and (b, 1). Find a and b.

3. We are given two points that are on the graph of y = arccos(x), which are (-1/2, a), and \((b, \frac{\pi}{4})\). Find a and b.

Thanks for the help!

Guest Sep 17, 2019

#1**+2 **

For 1:

\(a = -2\sin(\frac{-3\pi}{12})+1\) Now a can be calculated in a straightforward way.

and

\(1=-2\sin(3b)+1\) Now rearrange to find b.

I'm sure you can find a and b from these (note that there will be an infinite number of values of b).

Use the same approach for questions 2 and 3.

Alan Sep 17, 2019

#1**+2 **

Best Answer

For 1:

\(a = -2\sin(\frac{-3\pi}{12})+1\) Now a can be calculated in a straightforward way.

and

\(1=-2\sin(3b)+1\) Now rearrange to find b.

I'm sure you can find a and b from these (note that there will be an infinite number of values of b).

Use the same approach for questions 2 and 3.

Alan Sep 17, 2019