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anonymous
 5 years ago
sin inverse[sin 6pi/7]. I'm suppose to find the exact value.
anonymous
 5 years ago
sin inverse[sin 6pi/7]. I'm suppose to find the exact value.

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anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0that is true for all inverse functions f inverse f(x)=x

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0so the same thing applies for this: cos inverse[cos pi/2] is the the exact value pi/2 ?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0this is what inverse functions do

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0what if its negative like in this case: tan inverse[tan(pi/4)] ?

watchmath
 5 years ago
Best ResponseYou've already chosen the best response.2You need to be careful here, the inverse sine function has range (pi/2,pi/2). So the answer can't be 6pi/7.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0would you take the absolute value of "(pi/4)" so it would be "pi/4"

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0watchmath, i'm confused.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0oh true, did not think of that

watchmath
 5 years ago
Best ResponseYou've already chosen the best response.2What I am trying to say that \(\sin^{1}(\sin x)=x\) only for \(x\in[\pi/2,\pi/2]\)

watchmath
 5 years ago
Best ResponseYou've already chosen the best response.2So first you need to find an angle x so that sin(x)=sin(6pi/7). And we know that sin(pi/7)=sin(6pi/7) So \(\sin^{1}(\sin 6\pi/7)=\sin^{1}(\sin (\pi/7))=\pi/7.\)

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0so the exact value is pi/7?

watchmath
 5 years ago
Best ResponseYou've already chosen the best response.2Yes For the tangent, we have \[\tan^{1}(\tan x)=x\]only for \(x\in (\pi/2,\pi/2)\)
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