#sin^(-1)(4/3)#?

2 Answers
Harish Chandra Rajpoot
Jul 16, 2018

Undefined

Explanation:

Notice, that #\sin^{-1}(x)# is defined for any real number #x# such that it lies between #-1# & #1# including the limits i.e.

#\sin^{-1}(x)\ \ \text{defined for }\ \ -1\lex\le 1#

Since #4/3>1# hence, #\sin^{-1}(4/3)# is undefined.

Tony B
Jul 16, 2018

See explanation.

As written it is undefined.

Explanation:

Sine of an angle is #("opposite")/("hypotenuse") # for a right triangle.

For this triangle to exist the hypotenuse length must be greater that that of the opposite. The way you have written the question (if correct) means that the opposite is greater than the hypotenuse. Thus this triangle can not exist. It is impossible. So #sin^(-1)(4/3)# is'undefined'.

On the other hand; if you meant #sin^(-1)(3/4)# then this does exist thus it is 'defined'.

If you meant #->sin^(-1)(3/4) -> ubrace(arcsin(3/4))~~48.590377....#
#color(white)("dddddddddddddddddddddd")#I prefer this notation

#arcsin(3/4)~~48.59# to 2 decimal places

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