Expert: Josh - 2/17/2010

Question
Hi Josh, I really need your help on this question.
Find all the angles between -180<x<180 for which cos (2x-25) = 0.7
I am ok with questions like 0<x<180, but I am lost with negative degrees. Can help to solve and explain?
Thanks so much!

leo

Answer
Hi Leo,

Remember that sin(x), cos(x) and tan(x) are 2*pi periodic functions. If angle is measured in radian, the waveform repeats every 2*pi. If angle is measured in degrees, the values of the waveform repeats every 360 degrees.

Next thing to note is that all equations of the form sin(x)=A, cos(x)=A etc. (for some constant A) have two solutions in the interval from 0 to 360 degrees.

If we measure angles in the anti-clockwise direction from the right axis on an X-Y plane, -180 degrees coincides with the negative x-axis. Sweeping in an anti-clockwise direction, we reach the negative y-axis when angle is -90 degrees, as we increase the angle, at 0 degree we coincide with the positive x-axis. Keep going, we reach 90 degrees (effectively the positive y-axis) and eventually, we complete one revolution of the circle and we reach +180 degrees. We can cover the circle in different ways depending on our starting point. For example, -180 < x <= 180 and 0 <= x < 360 both cover one revolution (i.e., one period of the sin(x), cos(x) waveforms).

Remember this saying: "ASTC" (All Stations To Central).
This is used to remember the sign of each entity in each angle quadrant. viz.,

In the 1st quadrant: For 0< x <=90,
sin(x), cos(x) and tan(x) are always positive (non-negative).

In the 2nd quadrant: For 90< x <=180,
Only sin(x) is positive, cos(x) and tan(x) are both negative.

In the 3rd quadrant: For 180< x <=270 [equiv. to -180< x <=-90]:
Only tan(x) is positive, sin(x) and cos(x) are both negative.

In the 4th quadrant: For 270< x <=360 [equiv. to -90< x <=0]:
Only cos(x) is positive, sin(x) and tan(x) are both negative.

Hence, once we obtained a solution in the 1st quadrant, using the rules above,
we can determine the second solution in the fundamental interval -180 < x <= 180.
If x is between 0 and 90, the angles in other quadrants (2nd,3rd,4th) can be expressed as
2nd quad: 180-x (degree) or pi-x (radian)
3rd quad: 180+x (degree) or pi+x (radian)
4th quad: 360-x (degree) or 2*pi-x (radian)

e.g., consider sin(x)=1/2. In 1st quad, x=pi/6 or 30 degrees. Since sin(x) is +'ve in both the 1st and 2nd quadrants, the second solution comes from 180-x = 150 degrees.

e.g., for cos(x)=1/2, solution in 1st quad is x=pi/3 or 60 degrees. Since cos(x) is +'ve in both the 1st and 4th quadrants, the other solution comes from 360-x = 300 degrees. As the function has a period of 360 degrees, we can add or subtract integer multiple of 360 from this angle without any impact. In this case, we only need to subtract 300 by 360 to obtain -60 degrees. [Note: we expected this as cos(x) is an even function, symmetrical about the y-axis).

Returning to your original question.
cos (2x-25) = 0.7
2x-25 = cos^-1(0.7)
     = 45.572996, 360-45.572996 ...using ASTC, cos(x)>0 in 1st and 4th quadrant.

x = (45.572996+25)/2, (360-45.572996+25)/2
 = 35.28, 169.71 degrees approx.