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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!

 Sep 17, 2019

Best Answer 

 #1
avatar+28357 
+3

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.

 Sep 17, 2019
 #1
avatar+28357 
+3
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

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