Question 1116894
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We want equations in the form<br>
{{{a*sin(b(x-c))+d}}}<br>
and<br>
{{{a*cos(b(x-c))+d}}}<br>
The function has an amplitude of 3; so a in both equations is 3.<br>
The function has a maximum value of 4.  An amplitude of 3 and a maximum value of 4 means the minimum value is -2, and the centerline (d in each equation) 1.<br>
The period is 180 degrees, which is half the period of the basic sine or cosine function.  That means b is 2.<br>
Those are the relatively easy parts of the problem.  We have as the two equations<br>
{{{3*sin(2(x-c))+1}}}  and  {{{3*cos(2(x-c))+1}}}<br>
By far the hardest part (for most students) is finding the values for c in each equation.  That value determines the phase (horizontal) shift for the function.<br>
The given function has a maximum at x=0.  The basic cosine function has a maximum value at x=0, so there is no phase shift.  So the cosine equation for your example is<br>
{{{3*cos(2x)+1}}}<br>
For the sine function, the maximum occurs 1/4 of the way through the period; since the period of the function is 180 degrees (pi radians), the phase shift is 45 degrees (pi/4 radians) to the left.  So the phase shift c is -pi/4 (making "x-c" equal to "x+pi/4"), and the sine function is<br>
{{{3*sin(2(x+pi/4))+1}}}<br>
or<br>
{{{3*sin(2x+pi/2)+1}}}<br>
The first form is more meaningful to me, because I can see the phase shift of pi/4 clearly.