Question 1099401: Hi,
Please help me solve this non-calculator question.
First the question asks you to sketch a graph of y= sin( x ) for 0 ≤ x ≥ 360 which I've done already. It then says:
 = 30 degrees
Solve the equation  for 0 ≤ x ≥ 360.
How would I solve this using the fact that = 30 degrees?
Answer by Theo(13342)   (Show Source):
You can put this solution on YOUR website! 30 degrees is in the first quadrant of the unit circle.
the equivalent angle would be:
180 - 30 = 150 degrees in the second quadrant.
180 + 30 = 210 degrees in the third quadrant.
360 - 30 = 330 degrees in the fourth quadrant.
the trig functions for all of these angles would be the same except for the sign.
that's why they're called equivalent angles.
the signs of the trig functions are as shown below:
in the first quadrant, sine is positive, cosine is positive, tangent is positive.
in the second quadrant, sine is positive, cosine is negative, tangent is negative.
in the third quadrant, sine is negative, cosine is negative, tangent is positive.
in the fourth quadrant, sine is negative, cosine is positive, tangent is negative.
the sign of cosecant follows the sign of sine.
the sign of secant follows the sign of cosine.
the sign of cotangent follows the sign of tangent.
you are asked to find the value of x when sin(x) = -1/2 for 0 <= x <= 360.
since sine is negative in the third and fourth quadrant, that's where your angle will be.
the equivalent angle for 30 degrees in the third quadrant is 180 + 30 = 210 degrees.
the equivalent angle for 30 degrees in the fourth quadrant is 360 - 30 = 330 degrees.
you can use your calculator to confirm.
sin(30) = .5
sin(150) = .5
sin(210) = -.5
sin(330) = -.5
you can also graph the sine function and find the intersection of that graph with y = .5.
that will look like this.
quadrant 1 goes from 0 to 90
quadrant 2 goes from 90 to 180
quadrant 3 goes from 180 to 270
quadrant 4 goes from 270 to 360
|
|
|
