The concern is provided in the chap. 1.4 that the publication titled: Calculus troubles for the new century, by Robert Fraga.

You are watching: Sin(arcsec(x))

What is the domain the \$operatornamearcsec(sin x)\$?

I feel puzzled as \$operatornamearcsec( x ) = cos( x )\$, as \$sec( x) =arccos( x )\$. I hope the am not be make error in regards to domain values here, as domain of \$cos( x) \$ is entire real number line; while that of \$sec( x )\$ is in the term \$-1,cdots 1\$.

So, the concern reduces to finding domain the \$cos(sin x)\$.

The domain of \$cos(x)\$ is the selection of \$sin(x)\$, therefore domain of input to \$cos(x)\$, is \$-1le xle 1\$.

But, the book states answer as below, in an unanticipated way:

The \$operatornamearcsec(x)\$ is identified only ~ above \$\,\$. However, \$|sin, x,|ge 1\$ just if \$x\$ is an odd lot of of \$fracpi2\$, therefore the domain consists of strange multiples of \$fracpi2\$.

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edited Jul 10 "19 in ~ 12:34

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jitenjiten
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## 3 answer 3

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\$\$operatornamearcsec(x) extor sec^-1(x) e cos(x)\$\$

That declare is wrong. When we create something prefer \$sec^-1x\$, the does not typical \$frac1secx\$. It way "inverse secant function". It"s simply a notation thing. I know it"s confusion, but that"s just exactly how things room in modern-day inter-base.netematics.

As for the concern itself, the range of the role \$sinx\$ is \$<-1,1>\$. The train station secant function, \$operatornamearcsec(x)\$ (many times created as \$sec^-1x\$), is only identified on \$(-infty,-1>cup<1,+infty)\$. The intersection the those two sets is composed of the following two elements: \$-1,1\$. So, the domain that \$f(x)=sec^-1(sinx)\$ have to be all worths of \$x\$ whereby \$sinx\$ equals either \$-1\$ or \$1\$:

\$\$igg\fracpi2+kpi, kininter-base.netbbZigg.\$\$

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edited Jul 19 "19 at 17:06
reply Jul 10 "19 in ~ 11:48

Michael RybkinMichael Rybkin
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The arcsec is no the mutual of the \$sec\$ function. Examine the an interpretation of arcsec.

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reply Jul 10 "19 in ~ 11:48

Steve KassSteve Kass
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You mix up the inverse and also the reciprocal function: \$sec(x) = frac1cos(x) e arccos(x)\$.

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answer Jul 10 "19 at 11:55

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