# Thread: Moving a Trigonometric function

1. ## Moving a Trigonometric function

y=d+sin(bx+c)

So if I have the equation:
sin(2x)=y; I know that the period will be pi. So the whole graph will occur between (0,0) and (pi,0)
Start at (0,0), Maximum (pi/4, 1); intercept at (pi/2, 0); Minimum (3Pi/4, -1); Intercept (pi,0)

Now if I have:
y=sin(x-(pi/4)) the graph gets shifted to the right pi/4 radians. So key points are
Start: (pi/4, 0); Maximum: (3pi/4, 1); Intercept: (5pi/4,0); minimum (7pi/4, -1); Complete (9pi/4,0)

If I put them together as Sin(2x-(pi/4)
Why does my horizontal shift get cut in half? Why does it not start at (pi/4)? Why does it start at (pi/8)? In my graphing calculator I see everything gets moved my (pi/8) when I have this equation.

2. Originally Posted by IDontunderstand
y=d+sin(bx+c)

So if I have the equation:
sin(2x)=y; I know that the period will be pi. So the whole graph will occur between (0,0) and (pi,0)
Start at (0,0), Maximum (pi/4, 1); intercept at (pi/2, 0); Minimum (3Pi/4, -1); Intercept (pi,0)

Now if I have:
y=sin(x-(pi/4)) the graph gets shifted to the right pi/4 radians. So key points are
Start: (pi/4, 0); Maximum: (3pi/4, 1); Intercept: (5pi/4,0); minimum (7pi/4, -1); Complete (9pi/4,0)

If I put them together as Sin(2x-(pi/4)
Why does my horizontal shift get cut in half? Why does it not start at (pi/4)? Why does it start at (pi/8)? In my graphing calculator I see everything gets moved my (pi/8) when I have this equation.
because ... $\displaystyle y = \sin\left[2\left(x - \frac{\pi}{8}\right)\right]$

3. Thanks, that makes perfect sence. I did not see it that way. Thank you so much.