My Math Forum Finding exact value of the sum of two inverses

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 November 10th, 2018, 10:33 PM #1 Newbie   Joined: Nov 2018 From: Norway Posts: 2 Thanks: 0 Finding exact value of the sum of two inverses Hi, 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. Lars Last edited by skipjack; November 11th, 2018 at 02:08 AM.
 November 10th, 2018, 11:40 PM #2 Senior Member     Joined: Sep 2015 From: USA Posts: 2,408 Thanks: 1310 \begin{align*} &\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) = \\ &\dfrac{\sqrt{5}-4\sqrt{10}}{15} \end{align*} 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 $\sqrt{5}$ Thanks from larsh
 November 11th, 2018, 03:38 AM #3 Newbie   Joined: Nov 2018 From: Norway Posts: 2 Thanks: 0 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! https://image.ibb.co/mgBk3V/20181111-133353.jpg

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