Question 1038320
<pre><b>
Let the function be

{{{y = A*sin(bx+c)+d}}} 

[We use capital A here to avoid conflicting notation with
small "a" used here for some value that has nothing to do
with amplitude, which is usually |a| in other problems but 
here the amplitude will be |A|.  I have no idea why the 
author of your problem would have put in the "am" anyway,
since it just clutters the problem.  He should have given 
points (8,20) and (2,10).  But I'll clutter the problem with
the nonsensical "am", since he did.]

The function has max at point where {{{bx+c = pi/2}}} 
                                    {{{bx =pi/2-c}}}
                                    {{{x = (pi/2-c)/b}}}
                                    {{{x = (pi-2c)/(2b)}}}

Since we are given the max at (8am,20), that's where x = 8am,

So    {{{8am = (pi-2c)/(2b)}}}
      {{{16abm = pi-2c}}}
     
and where y = 20, that's where

{{{y = A*sin(bx+c)+d}}}
{{{20 = A*sin(pi/2)+d}}}
{{{20 = A*1+d}}}
{{{A+d=20}}}

========================

{{{y = A*sin(bx+c)+d}}} [Use capital A to avoid conflicting notation]

has min at point where {{{bx+c = 3pi/2}}} 
                       {{{bx =3pi/2-c}}}
                       {{{x = (3pi/2-c)/b}}}
                       {{{x = (3pi-2c)/(2b)}}}

Since given the min at (2am,10), that's where x = 2am,

So    {{{2am = (3pi-2c)/(2b)}}}
      {{{4abm = 3pi-2c}}}
     
and where y = 10, that's where

{{{y = A*sin(bx+c)+d}}}
{{{10 = A*sin(3pi/2)+d}}}
{{{10 = A*(-1)+d}}}
{{{A-d=10}}}

So we have the two systems of equations:

{{{system(16abm = pi-2c,4abm = 3pi-2c)}}} 

and

{{{system(A+d=20,A-d=10)}}}

From the first system:

{{{system(16abm = pi-2c,4abm = 3pi-2c)}}}

Multiply the second by -4

{{{system(16abm = pi-2c,-16abm = -12pi+8c)}}}

Adding the equations gives;

{{{0=4pi+6c}}}
{{{-6c=4pi}}}
{{{c=-4pi/6}}}
{{{c=-2pi/3}}}

Substituting in 4abm = 3pi-2c

{{{4abm = 3pi-2c}}}
{{{4abm = 3pi-2(2pi/3)}}}
{{{4abm = 3pi-4pi/3}}}
{{{12abm = 9pi-4pi}}}
{{{12abm=5pi}}}
{{{b=5pi/(12am)}}}

From the second system:

{{{system(A+d=20,A-d=10)}}}

Adding them gives

{{{2A=30}}}
{{{A=15}}}

Subtracting them gives"

{{{2d=10}}}
{{{d=5}}}

So the equation

{{{y = A*sin(bx+c)+d}}}

becomes

{{{y = 15*(expr(sin(5pi/(12am))x+expr(2pi/3)^""))+5}}}

Edwin</pre></b>