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# questions

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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

#1
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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
+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.

Alan Sep 17, 2019