8
$\begingroup$

Does $\operatorname{arcsec}(x) = 1 /\arccos(x)$? I have looked in a few books and Google'd it but I am not finding my answer.

$\endgroup$
4

8 Answers 8

29
$\begingroup$

If $\sec^{-1} x = \theta$, then $x = \sec\theta$. This means $\frac1x = \cos\theta$, so $\cos^{-1}\frac1x = \theta$. So your equation is wrong; the correct statement is $$\boxed{\sec^{-1} x = \cos^{-1}\tfrac1x}$$

$\endgroup$
1
  • $\begingroup$ $$(\sec x)^{-1} = (\cos x)^1 \\ \sec^{-1} x = \cos^{-1} x^{-1}$$ it’s kinda satisfying . . . feels like the negative ones cancel each other out $\endgroup$ Apr 26, 2019 at 21:18
14
$\begingroup$

Actually it's: $$\operatorname{arcsec}(x)=\arccos(1/x).$$

$\endgroup$
5
$\begingroup$

No. It is not. If you look at the definitions

$$y=\frac{1}{\cos x}$$

and then we solve for the x

$$\frac{1}{y}=\cos x$$

$$\cos^{-1}\left(\frac{1}{y}\right) = x$$

and replace $x$ and $y$ to find the inverse

$$y=\cos^{-1}\left(\frac{1}{x}\right)$$

$\endgroup$
1
  • $\begingroup$ Careful, $\cos^{-1}(\cos(x)) \neq x$ in some cases. $\endgroup$ May 17, 2014 at 18:49
5
$\begingroup$

It is straight forward, let $$\sec^{-1}(x)=\alpha$$ $$\implies \sec\alpha=x$$ $$\implies \frac{1}{\cos \alpha}=x$$$$\implies \cos \alpha=\frac{1}{x}$$ $$\implies \alpha=\cos^{-1}\left(\frac{1}{x}\right)$$ Substituting the value $\alpha=sec^{-1}(x)$, we get $$\bbox[4pt, border:1px solid blue;]{\color{red}{\sec^{-1}(x)=\cos^{-1}\left(\frac{1}{x}\right)}}$$

$\endgroup$
4
$\begingroup$

No. If you graph $\sec^{-1}(x) \cdot \cos^{-1}(x)$, you get:

-2 to 2 -15 to 15

You can clearly see that it isn't $1$.

$\endgroup$
3
$\begingroup$

No, it is false. Probably you meant $\operatorname{arcsec}(x)=\arccos(1/x)$, which is true.

$\endgroup$
2
$\begingroup$

Isn't. Draw $$\text{arcsec} x\arccos x$$

$\endgroup$
1
  • 2
    $\begingroup$ Please avoid link only answers and add a plot if you are telling the OP to test his theory by graphing. $\endgroup$
    – Cole Tobin
    May 16, 2014 at 17:03
1
$\begingroup$

No, they are not equal. Here is why.

We can look at an algebraic approach to arcsec and arccos first:

Arccos(x) = y
Cos(y) = x
1/sec(y) = x
Sec(y) = 1/x
y = arcsec(1/x)

Now, we have established that Arccos(x) = y = arcsec(1/x), Thus, arccos(x) = arcsec(1/x) And we have got a neat algebraic proof.

Conversely, we can do the reverse by substituting cos for sec in the above equations to get arcsec(x) = arccos(1/x)

Take a graphing calculator, you can substitute the values, and this is the image of the graph

Lastly, the longest side of a triangle is 1/cos(x) = y, and since y is not equal to sec(y), the above equation in the question is wrong ( 1/cos(x) is not sec(y), and 1/arccos(x) is not arcsec(x) ).

$\endgroup$

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .