Question

Trig identities:
finding the exact value of expression. if the expression is not valid, state why it is invalid. without using calculator

sin[sin^1(2)]

Answer 1

im sure what the question is asking but the answer to the problem itself is 0.7891 i believe

Answer 2

not sure*

Answer 3

thats what i think

Answer 4

sorry man

Answer 5

:(

Answer 6

Sorry i tried

Answer 7

thanks anyways :)

Answer 8

you mean \[\sin (\sin2) ?\]

Answer 9

no it is a trick question

Answer 10

range of sine is
\[[-1,1]\] so domain of arcsine is
\[[-1,1]\] and therefore you cannot take the arcsine of 2 since it is not in the domain

Answer 11

in other words there is no such number x with
\[\sin(x)=2\] and therefore
\[\sin^{-1}(2)\] does not exist

Answer 12

no such number

Answer 13

since
\[\sin^{-1}(2)\] does not exist, neither does
\[\sin(\sin^{-1}(2))\]
for that matter neither does
\[(\sin^{-1}(2))^2\] or any other expression containing
\[\sin^{-1}(2)\]

Answer 14

they are trying to trick you in to saying
\[\sin(\sin^{-1}(2))=2\] reasoning that
\[f(f^{-1}(x))=x\] but don't fall for the trap

Answer 15

:D thank you..i have more question by the way..which ask the same valid or invalid query

Answer 16

cos^-1[cos (7pi/3)]

Answer 17

since \(\cos(\frac{7\pi}6)=-\frac12\) you should be able to answer the question with what satellite has said

Answer 18

actually \(\cos(\frac{7\pi}6)=-\frac{\sqrt3}2\), but same answer...

Answer 19

7pi/3 ..not 7pi/6 @TuringTest

Answer 20

whatever, you are missing the point:

\(\sin\theta,\cos\theta\le1\) for all possible \(\theta\)
therefor \(\cos(\frac{7\pi}3)<1\)

(it happens that \(\cos(\frac{7\pi}3)=\frac12\), but all that matters is that it is less than one)

the idea here being that they want you to recognize that \(\sin^{-1}x\) and \(\cos^{-1}x\) are undefined for \(x>1\)

so as long as that does not happen, the expression has meaning;

since \(\cos\theta\le1\) then \(\cos^{-1}(\cos\theta)\) is defined, and so in this case we do in fact have \(\cos^{-1}(\cos\theta)=\theta\)

Answer 21

I trust those answers provided by TuringTest and satellite73. Stay within the domain of those functions.

Answer 22

@gbluedinosaur, unable to respond via the message feature.