Question

sin^(-1)(sin(7pi/6))

Answer 1

hint: (this is a usual scenario but does not happen all the time...but most of the questions in schools follow this principle)

\[\huge \sin (\sin^{-1} (x)) = x\]
\[\huge \sin^{-1} (\sin (x) ) = x\]
does that help?

Answer 2

same thin with
\[\huge \cos (\cos^{-1} (x)) = x\]
\[\huge \cos^{-1} (\cos (x)) = x\]

Answer 3

or \[\huge \tan (\tan^{-1} (x)) = x\]
\[\huge \[\tan^{-1} (\tan (x)) = x\]

Answer 4

i really hope im making sense here =))

Answer 5

ummm..hold on..

Answer 6

sure

Answer 7

so does x just equal 7pi/6?

Answer 8

yup

Answer 9

GREAT! thanks once again!

Answer 10

welcome ^_^

Answer 11

what if its like this: cos(sin^(-1)(x)?

Answer 12

draw a triangle

Answer 13

sin^(-1)(x) is the same as sin(theta)= x/1

Answer 14

so find the missing side using pythagorean's theorem, then find cos(theta),


if im wrong, someone please correct me
i havent done a problem like this in a while

Answer 15

here, i'm gonna start a new post with the exact problem.