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# Find phase when moving a sine graph

Question

The graph of f(x) = sin 3x + cos 3x move right C units, the result graph is symmetry to the y-axis. Find the minimum value of C.

Solution

f(x) = sin3x + cos3x
= (sin3x + cos3x) (2/2)
= (sin3x + cos3x) Sqrt (2) × Sqrt (2)/2
= Sqrt (2)[sin3x Sqrt (2)/2 + cos3x Sqrt (2)/2]
= Sqrt (2) (sin3x cos pi/4 + cos3x sin pi/4)
= Sqrt (2) sin(3x + pi/4)

We need to find the equation of those vertical lines that the graph of f(x) = sin (3x + pi/4) will symmetry to.

Now we think about the graph of y = sin x. when x is range from 0 – 2pi, the graph of y = sin x is symmetry to the vertical line x = pi/2 and x = 3pi/2. The distance between the adjacent vertical symmetry line is pi, so the equation of the symmetry line of y = sin x is: x = pi/2 + pi K, in which K = 0, +-1, +-2, …

To find the symmetry axis of y = sin (3x + pi/4), let 3x + pi/4 = pi/2 + pi K,

3x = - pi/4 + pi/2 + pi K
3x = -pi/4 + 2pi/4 + pi K
3x = pi/4 + pi K
x = pi/12 + (pi/3) K

Because we are asked to move the graph right C units to let it symmetry to the y-axis, so Let K = -1,

When K = -1,
x = pi/12 + pi/3 × (-1)
= pi/12 - pi/3
= pi/12 – 4pi/12
= -3pi/12
= -pi/4

Therefore, move the graph of f(x) = sin 3x + cos 3x right pi/4 units, the result graph will symmetry to the y-axis. So, C = pi/4. Look the graph of f(x) = sin3x + cos3x above, the graph of f(x) is symmetry to the line x = - pi/4. When move the graph right pi/4 units, the result graph will symmetry to the y-axis. Watch the video for more details.