Question 1205013
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Many teachers and textbooks decided that sin x is fine when others prefer sin(x) to make it more clear what the input argument is. 
The second version leads to more parenthesis but it's worth it I think.


Something like  sin B(x + C) is the same as  sin( B(x + C) ) to tell the reader "The B(x+C) part is inside the sine function".
It's a bit confusing I know.


No, y = A sin B(x + C) + D is NOT the same as y = A*sin(Bx - C) + D; however, the two are somewhat similar at least.
The A, B, and D play the exact same role for either template. The C is a bit different. 


Here's an example graph showing that y = A sin B(x + C) + D is NOT the same as y = A*sin(Bx - C) + D
<a href="https://www.desmos.com/calculator/sbgketg7yy">https://www.desmos.com/calculator/sbgketg7yy</a>


Check out this page 
<a href="https://math.libretexts.org/Courses/Rio_Hondo/Math_175%3A_Plane_Trigonometry/02%3A_Graphing_Trigonometric_Functions/2.04%3A_Transformations_Sine_and_Cosine_Functions">https://math.libretexts.org/Courses/Rio_Hondo/Math_175%3A_Plane_Trigonometry/02%3A_Graphing_Trigonometric_Functions/2.04%3A_Transformations_Sine_and_Cosine_Functions</a>
to refresh your memory of what each parameter does.
The template that link uses is y = A*sin(Bx - C) + D


However, since your teacher is using the template 
y = A*sin(B(x + C)) + D
it's best to stick to it


|A| = amplitude
B = used to determine period
C = used to determine phase shift
D = vertical shift and midline


The highest and lowest points on this sinusoidal curve is when y = 1 and y = -3 respectively. 
The midpoint is (max+min)/2 = (1+(-3))/2 = -1 which is the midline. Therefore <font color=red>D = -1</font>


The vertical gap between those y values is 4 units. Count out the spaces. Or subtract and use absolute value. Half of this vertical distance is the amplitude. So either A = 2 or A = -2. Both will work. I'll go with <font color=red>A = 2</font> for simplicity sake, and to match what your answer key shows (note: if you picked A = -2, then the C value would be different from C = -pi/3 while everything else is the same).


The lowest point occurs when y = -3
Circle two neighboring valley points and record their x values. Two such x values are x = pi/6 and x = 5pi/6
The horizontal gap between them is 5pi/6 - pi/6 = 4pi/6 = 2pi/3
This is the period. It's the length of each cycle. The curve repeats itself every 2pi/3 horizontal x units.


T = period
T = 2pi/B
B = 2pi/T
B = 2pi/( 2pi/3 )
B = 2pi * (3/(2pi))
<font color=red>B = 3</font>
Note how,
T = 2pi/B
T = 2pi/3
So we confirm we have the correct B value.


The last thing to find is the value of C.
So far we have:
<font color=red>A = 2</font>
<font color=red>B = 3</font>
<font color=red>D = -1</font>
Plug those into the template y = A*sin( B(x + C) ) + D and we get y = 2*sin( 3(x + C) ) - 1
Then plug in one of the min or max points on the curve. Let's say we picked (x,y) = (pi/2, 1)


Let's solve for C.
y = 2*sin( 3(x + C) ) - 1
1 = 2*sin( 3(pi/2 + C) ) - 1
1 + 1 = 2*sin( 3(pi/2 + C) )
2 = 2*sin( 3(pi/2 + C) )
2/2 = sin( 3(pi/2 + C) )
1 = sin( 3(pi/2 + C) )
3(pi/2 + C) = arcsin(1)
3(pi/2 + C) = pi/2
3pi/2 + 3C = pi/2
3C = pi/2 - 3pi/2
3C = -2pi/2
3C = -pi
<font color=red>C = -pi/3</font>


Here's a Desmos graph to confirm our answers
<a href="https://www.desmos.com/calculator/iosov8wrk1">https://www.desmos.com/calculator/iosov8wrk1</a>
GeoGebra is another graphing tool I use all the time.
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