How would you evaluate this expression?

Question

How would you evaluate this expression?

cos^(-1)(-squareroot3/2)

Answer 1

function Cos^ (-1)x is an inverse of function Cos x
(like square(x) and squareroot(x) act oppostite
square(sqrootx) = x or sqroot(square(x) = x
Therefore Cos[Cos^(-1)x = x
Now to your question
Let [Cos^(-squareroot3/2) = x
Taking Cos on both side
Cos[Cos^(-1)(-squareroot3/2)] = Cos x
or -squareroot3/2 = Cos x
This equation has many solutions as you can see (for every cycle of 360 deg you will have two solutions - one each in second quadrant and third quadrant where Cosine (or cos) has negative values
Primarily we take the solution which between zero and 360 degrees
Two solutions are
Cos (180 -30) = Cos (180+30) = (- squareroot 3/2)
therefore 150 deg and 210 deg
converted to 150xpi/180 = 5/6 pi radians
and 210xpi/180 = 7/6 pi radians
Similarly we can find a set of two solutions between 360 &720 deg, between 720 &1080 deg etc and so on
eg [360 + (180+30)] deg and [360 +180-30] deg [720 + (180+30)] deg [720+ (180-30)] deg and so on
Cheers
Subhash

Answer 2

cos^(-1) means the inverse of cosine, or OPPOSITE of cosine.

So cos^(-1)(-squareroot3/2) actually means:

what number do I take the cosine of in order to get an answer of (-squareroot3/2)?

Well, to get an answer of (-squareroot3/2), you could take the cosine of 5pi/6

You can also do this on your calcular using the cos^(-1) key. There should be a button that says cos^(-1), you usually have to push the "2nd" key to get to it.