Question

what is the principal value of cos^-1(-rad3/2)

Answer 1

Explanation please.

Answer 2

I too..

Ha ha ha..

Answer 3

\[\large \cos^{-1}(-\theta) = \pi - \cos^{-1}(\theta)\]

Answer 4

what are you trying to show me?

Answer 5

So you will get :
\[\cos^{-1}(-\frac{\sqrt{3}}{2}) = \pi - \cos^{-1}(\frac{\sqrt{3}}{2})\]

Answer 6

So now solve for:
\[\cos^{-1}(\frac{\sqrt{3}}{2}) \implies \cos^{-1}(\cos(\frac{\pi}{6}))\]

Answer 7

oh i got it now

Answer 8

Thank you :D

Answer 9

And you know:
\[\cos^{-1}(\cos(x)) = x\]

Use this and go further..

Answer 10

Do not forget to subtract it from \(\pi\)...

You are welcome dear..

Answer 11

If any doubt you can ask us anytime....

Answer 12

Thank you for saying that! I definitely will, once again thank you for the help.

Answer 13

You are welcome..