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zryanomer45 April 18th, 2018 06:09 AM

Pove
 
Hi how can prove cos-1(-x)=-cos-1x


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Country Boy April 18th, 2018 08:14 AM

If you cannot use Latex, at least use "^" to indicate the inverse function. You want to prove that cos^(-1)(x)= -cos^(-1)(-x) ($\displaystyle cos^{-1}(x)= -cos^{-1}(-x)$.

Your basic problem is that you can't prove that, it isn't true!

For example, if $\displaystyle x= \frac{\sqrt{2}}{2}$ then $\displaystyle cos^{-1}(x)= \frac{\pi}{4}$ radians (45 degrees). But the definition of the function $\displaystyle cos^{-1}(x)$ restricts the range to between 0 and $\displaystyle \pi$ (180 degrees). $\displaystyle cos^{-1}\left(-\frac{\sqrt{2}}{2}\right)= \frac{3\pi}{4}$ (135 degrees), NOT $\displaystyle -\frac{\pi}{4}$ radians or -45 degrees.

If you were to change "$\displaystyle cos^{-1}$" to "$\displaystyle sin^{-1}$" then it would be true.

skeeter April 18th, 2018 10:23 AM

Quote:

Originally Posted by zryanomer45 (Post 592717)
Hi how can prove cos-1(-x)=-cos-1x


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A correct equation that may be proven could be ...

$\cos^{-1}(-x) = \pi - \cos^{-1}(x)$

zryanomer45 April 18th, 2018 10:45 AM

Quote:

Originally Posted by skeeter (Post 592739)
A correct equation that may be proven could be ...



$\cos^{-1}(-x) = \pi - \cos^{-1}(x)$



Ok how can l prove it


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romsek April 18th, 2018 11:05 AM

$\cos^{-1}(-x) = -\cos^{-1}(x)$

$-x = cos(-\cos^{-1}(x))$

the cosine is an even function so $\cos(-x)=\cos(x)$ thus

$-x = \cos(\cos^{-1}(x)) = x$

and thus as Country Boy keenly noted your statement isn't true.

skeeter April 18th, 2018 11:10 AM

evaluate the cosine of each side ...

$\cos[\cos^{-1}(-x)] = -x$

$\cos[\pi - \cos^{-1}(x)] = \cos(\pi)\cos[\cos^{-1}(x)] + \sin(\pi)\sin[\cos^{-1}(x)] = (-1)(x) + 0 = -x$


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