Question

sec[inverse sin(-1/2)]
explain how to get the answer 2sqrt3/3

Answer 1

\[\sec [\sin^{-1} (-1/2)]=(2\sqrt{3})/(3)\]
someone explain to me how that answer is right..pleasee!

Answer 2

does it help to know that
\[\sin^{-1}(-\frac{1}{2})=-\frac{\pi}{6}\]?

Answer 3

i knew that but how in the world does the sec come out to that?

Answer 4

first you\[\sin^-1(-1/2)=-\frac{ \pi }{ 6 }\]
\[\sec(-\frac{ \pi }{ 6 })=\frac{ 1 }{ \cos-(\frac{ \pi }{ 6 } )}=\frac{ 1 }{ \frac{ \sqrt3 }{ 2 } }=\frac{ 2 }{ \sqrt3 }\]

Answer 5

bc the cos would be \[(\sqrt{3})/(2)\]

Answer 6

Secant is the reciprocal of Cosine, i.e.

\[\sec(\theta)=\frac{ 1 }{ \cos(\theta) }\]

Answer 7

Do you see now where I got 2/sqrt(3) from?

Answer 8

yea i get that sec is the reciprocal of cos so shouldnt the answer be 2/sqrt3