Theta in converting sec to cos

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SUMMARY

The discussion centers on the conversion of the inverse secant function to the inverse cosine function, specifically from arcsec(4/√π) to arccos(√π/4). The key takeaway is that since sec(θ) and cos(θ) are reciprocals, the conversion involves flipping the fraction. The solution provided confirms that if x = sec⁻¹(4/√π), then sec(x) = 4/√π leads to cos(x) = √π/4, establishing that x = cos⁻¹(√π/4). It is important to note the domain restrictions on x when dealing with inverse trigonometric functions.

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Homework Statement


I saw in my calculus book that something along the lines of arcsec [itex]\frac{4}{\sqrt{pi}}[/itex] = x was converted to arccos [itex]\frac{\sqrt{pi}}{4}[/itex] = 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 [itex]\frac{4}{\sqrt{pi}}[/itex] = x was converted to arccos [itex]\frac{\sqrt{pi}}{4}[/itex] = 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 = [itex]sec^{-1}\frac{4}{\sqrt{\pi}}[/itex]
Then sec(x) = [itex]\frac{4}{\sqrt{\pi}}[/itex]
So cos(x) = [itex]\frac{\sqrt{\pi}}{4}[/itex]
Which means that x = cos-1[itex]\frac{\sqrt{\pi}}{4}[/itex]

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

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