Theta in converting sec to cos

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The discussion centers on the conversion between arcsec and arccos functions, specifically how arcsec(4/√π) can be expressed as arccos(√π/4). The key point is that since sec and cos are reciprocal functions, the relationship necessitates flipping the values when converting between them. The participant expresses confusion about this transformation and seeks clarification on the reasoning behind it. A response explains the process by defining x as sec^(-1)(4/√π) and deriving that cos(x) equals √π/4, leading to x being equal to cos^(-1)(√π/4). The importance of understanding domain restrictions for x is also highlighted.
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Homework Statement


I saw in my calculus book that something along the lines of arcsec \frac{4}{\sqrt{pi}} = x was converted to arccos \frac{\sqrt{pi}}{4} = x. I understand that sec and cos are reciprocals, but I don't see why has to be flipped as well.

Homework Equations


cos (θ), sec (θ)


The Attempt at a Solution


I began to think of the graphs and such, but I just can't think of why this works. I don't normally mess with the inverse trig functions and was hoping someone could point out what I'm missing.

Thank you.
 
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vanmaiden said:

Homework Statement


I saw in my calculus book that something along the lines of arcsec \frac{4}{\sqrt{pi}} = x was converted to arccos \frac{\sqrt{pi}}{4} = x. I understand that sec and cos are reciprocals, but I don't see why has to be flipped as well.

Homework Equations


cos (θ), sec (θ)


The Attempt at a Solution


I began to think of the graphs and such, but I just can't think of why this works. I don't normally mess with the inverse trig functions and was hoping someone could point out what I'm missing.

Thank you.
Let x = sec^{-1}\frac{4}{\sqrt{\pi}}
Then sec(x) = \frac{4}{\sqrt{\pi}}
So cos(x) = \frac{\sqrt{\pi}}{4}
Which means that x = cos-1\frac{\sqrt{\pi}}{4}

It should be understood that there are domain restrictions on x.
 
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