back to Trigonometry video lessons

# Find the analytical expression of the sine graph

Question:

The figure shows a part of the sine graph y = sin (Bx + C), find the analytical expression of the sine graph.

Solution:

The graph is a part of the sine wave. To find the analytical expression of the sine wave, we measured the coordinates of points A and B. The x-coordinate of the point A is -2pi/5, the y-coordinate of the point A is 0. The x-coordinate of the point B is -3pi/20, the y-coordinate of the point B is the highest value 1. So, we know the amplitude of the sine wave is 1.

The value of y = sin (Bx + C) rises from zero to the highest point after a quarter cycle. Let T be the minimum period, then T/4 = -3pi/20 - (-2pi/5) = -3pi/20 + 2pi/5 = -3pi/20 + 8pi/20 = (8pi - 3pi)/20 = 5pi/20 = pi/4. So, T = pi

From T = 2pi/B, then B = 2pi/T = 2pi/pi = 2

So, the sine function is: y = sin (2x + C)

Now, we will determine the initial phase C

Remember the graph of y = sin x, when x = 0, y = 0.

For the graph of y = sin (2x + C), when 2x + C = 0, y = 0, which corresponding to point A, where x = - 2pi/5

So, 2x + C = 0, where x = - 2pi/5
2 × (-2pi/5) + C = 0
-4pi/5 + C = 0
both sides of the equation plus 4pi/5
C = 4pi/5
therefore, the initial phase C is 4pi/5

The analytical expression of the sine wave is y = sin (2x + 4pi/5). Watch the video for more details.