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  • 1 Post By romsek
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November 10th, 2018, 10:33 PM   #1
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Finding exact value of the sum of two inverses


From my textbook, I have an example saying:

cos(arcsec(3) + arctan(2)) = (√5 − 4√10)/15

Why is it so? If anyone would want to explain, I would be very grateful.

If this had been angles, I could recognize, ex: cos(arcsec(2) + arctan(1)) I would have used the sum identity for cos, so it would be cos(pi/3 + pi/4), It's especially this arcsec(3) which bothers me.


Last edited by skipjack; November 11th, 2018 at 02:08 AM.
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November 10th, 2018, 11:40 PM   #2
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&\cos(\sec^{-1}(3)+\arctan(2)) = \\

&\cos(\sec^{-1}(3))\cos(\arctan(2)) - \sin(\sec^{-1}(3))\sin(\arctan(2)) = \\

&\dfrac 1 3 \dfrac{1}{\sqrt{5}} - \dfrac{2\sqrt{2}}{3}\dfrac{2}{\sqrt{5}} = \\

&\dfrac{1}{3\sqrt{5}}\left(1-4\sqrt{2}\right) = \\



that $\cos(\sec^{-1}(3)) = \dfrac 1 3$ should be fairly obvious

$\sin(\sec^{-1}(3)) = \dfrac{2\sqrt{2}}{3}$ comes from $\sin(x) = \sqrt{1-\cos^2(x)}$

to see $\cos(\arctan(2))$ and $\sin(\arctan(2))$ consider a right triangle with opposite side of length 2, adjacent side of length 1, and hypotenuse of length $\sqrt{5}$
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November 11th, 2018, 03:38 AM   #3
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Oh, thanks. That made a lot of sense I think I sometimes don't see the forest for all the trees.

It totally made sense and with your guidance I was able to reproduce the solution and understand everything. One million thanks's!
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