Question

sin^-1(-3/5)

Answer 1

what are you supposed to do with it?
if you have to compute it, you need a calculator

Answer 2

step by step calculation

Answer 3

well there are several methods.
are you aware you can move inverse sin to the other side by inverse operations and evaluate it that way?

Answer 4

there is nothing you can do but use a calculator, unless this is part of some other problem

Answer 5

basically what he said. the information is useless without direction

Answer 6

for example, if it is
\[\cos(\sin^{-1}(-\frac{3}{5}))\] then you can compute something

Answer 7

but
\[\sin^{-1}(-\frac{3}{5})\] is just a number that you will not know without a calculator, there is no step by step way to find it

Answer 8

satellite you are right the full calculation in for cos(sin^-1(-3/5)). my question is why sin^-1(-3/5) when calculated by pythagoras gives cos^-1(4/5) and not cos^-1(-4/5)

Answer 9

how's i guess?

Answer 10

very good

Answer 11

think about the quadrant signs

Answer 12

there is a picture of an angle where the sine is \(\frac{3}{5}\)
by pythagoras (or by memory of the mother of all right triangles) you know the other side is \(4\)

Answer 13

lonnie455rich quad can give both +4/5 or -4/5 depending in which quad we make our triangle

Answer 14

then the cosine is pretty obvious

Answer 15

well if your problem shows that cos is positive and sin is negative. you know what quadrant it is. or were you asking which one to pick?

Answer 16

no need

Answer 17

it is positive for sure

Answer 18

i can give you at least two reasons why it must be positive, one is conceptual and the other is a bit fancier, but amounts to the same thing

Answer 19

no satellite my question is different. if triangle is in third quad as per question then the answer is cos^-1(-4/5), and the answer will be cos(cos^-1(-4/5)) = -4/5

Answer 20

lets go slow

Answer 21

yes why should it be positive always

Answer 22

\[\cos(\sin^{-1}(-\frac{3}{5}))\] this is the actual question right, not some other question

Answer 23

abs correct

Answer 24

ok fancy reason first

Answer 25

go ahead pls

Answer 26

he is typing.

Answer 27

sine is odd, and therefore \(\arcsin\) is odd as well
that means
\[\sin^{-1}(-\frac{3}{5})=-\sin^{-1}(\frac{3}{5})\] and cosine is even making
\[\cos(\sin^{-1}(-\frac{3}{5}))=\cos(-\sin^{-1}(\frac{3}{5}))=\cos(\sin^{-1}(\frac{3}{5}))\]

Answer 28

ok

Answer 29

that is the fancy reason, here is the plain reason
the range of \(\arcsin\) is \([-\frac{\pi}{2},\frac{\pi}{2}]\) in other words only in quadrant 1 and 4
in those quadrants cosine is positive

Answer 30

niiiice. i didnt think to consider that.

Answer 31

understood.

Answer 32

in other words,
\[\sin^{-1}(-\frac{3}{5})\] is not SOME number whose sine is \(-\frac{3}{5}\) but only the unique number between \(\frac{\pi}{2}\) and \(-\frac{\pi}{2}\) whose sine is \(-\frac{3}{5}\) and so the cosine must be positive

Answer 33

satellite thanks a great deal pal

Answer 34

yw