We have:
cos−1(−12)=x
Rearrange:
−12=cos(x)
Now we ask, which values of x would produce −12 by the cosine function? This is possible because −12 is a value we encounter regularly with the cosine and sine functions. We know that a right triangle with angle 60o adjacent to the horizontal side will give us a cosine of 12. In the correct quadrant, namely quadrant II, we can get −12.
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Where a=1.
The cosine in this case is given by xr or the ratio of the adjacent side over the hypotenuse. This gives cos(60o)=12.
In the second quadrant, x<0 and y>0, giving cos(120o)=−12 (we use 120o because it gives us an angle of 60o above the negative x-axis).
Quadrant II:
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The radian equivalent of 120o is 2π3. Because the cosine function repeats every 2π units (i.e. has period of 2π), you may also write arccos(−12)=2π3±2π.
