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# How to find the set of angles when given the value of sin x?

Question:

Given: sin x = - 0.7660, in which x is from 0 to 2pi. Find the set of the angle x.

Solution:

Because sin x defined as, sin x = y/r, in which r = square root of x square plus y square. So, r is always a positive number. When y is a negative number, sin x is a negative number. There will be an angle x that will satisfy sin x = - 0.7660.

Let x1 be an angle in Quadrant III. The terminal side of the angle x1 lies in Quadrant III. P is a point on the terminal side of the angle x1. P has coordinate (x, y). Then sin x1 = y/r, since y in Quadrant III is a negative number, then sin x1 is a negative number. So, there is an angle x1 that will satisfy sin x = - 0.7660.

Let the angle x2 be an angle in Quadrant IV. The terminal side of the angle x2 lies in Quadrant IV. P is a point on the terminal side of the angle x2. P has coordinate (x, y). Then sin x2 = y/r. Because y in Quadrant IV is a negative number, then sin x2 is a negative number. There is an angle x2 that will satisfy the equation sin x = - 0.7660.

Now we will find an acute angle that corresponding to the absolute value of the given trigonometry function. That is, let sin x = |-0.7660| = 0.7660. Then x = arc sin 0.7660 = 50o. Then the angle in Quadrant III is: x1 = 180o + 50o = 230o. The angle in Quadrant IV is: x2 = 360o - 50o = 310o. Therefore, the set of angles that satisfy sin x = - 0.7660 and the angle x range from 0 to 2pi is {230o, 310o}. Look the figure above, the blue curve is the graph of y = sin x. The orange line is y = - 0.7660. The intersection points of the blue curve and orange line is x1 and x2. So, x1 and x2 are the solutions that satisfy the given equation sin x = - 0.7660 and the angle x range from 0 to 2pi. x1 lies in Quadrant III and x2 lies in Quadrant IV. The angle set is {230o, 310o}. Watch the video for more details.