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+0  
 
0
149
4
avatar+895 

Disregard, I figured it out.

 

Find the exact value:

 May 11, 2019
edited by AdamTaurus  May 11, 2019
 #1
avatar+7712 
+1

\(\cos^{-1}(1) = 0\\ \tan\left(\sin^{-1}\left(\dfrac{4}{5}\right)+\cos^{-1}(1)\right)\\ = \tan\left(\sin^{-1}\left(\dfrac{4}{5}\right)\right)\\ = \dfrac{4}{\sqrt{5^2-4^2}}\\ =\dfrac{4}{3}\)

.
 May 11, 2019
 #2
avatar+7712 
0

\(\tan\left(\cos^{-1}\left(\dfrac{4}5{}\right)+\sin^{-1}\left(1\right)\right)\\ =\tan\left(\dfrac{\pi}{2}+\cos^{-1}\left(\dfrac{4}5\right)\right)\\ =\dfrac{1}{\tan\left(\cos^{-1}\left(\dfrac{4}5\right)\right)}\\ =\dfrac 1{\dfrac{\sqrt{5^2-4^2}}4}\\ =\dfrac{4}{3}\)

.
 May 11, 2019
edited by MaxWong  May 11, 2019
 #3
avatar+8756 
+3

Adding  \(\frac{\pi}{2}\)  flips the tangent and negates it. If     \(\tan\theta\,=\,\frac{a}{b}\)     then    \(\tan(\theta+\frac{\pi}{2})\,=\,-\frac{b}{a}\)    

hectictar  May 11, 2019
 #4
avatar+103917 
+1

tan ( arccos(4/5) + arcsin(1) ) =

 

tan (arcos(4/5) + pi/2)

 

Let    arccos(4/5)   =  θ           so  cos θ = 4/5       and sin θ = 3/5

 

So we have

 

tan ( θ + pi/2)  =

 

sin ( θ + pi/2)               sinθ cos pi/2   +  sin pi/2 * cos θ                 cos θ

___________   =        ___________________________   =    __________ =

cos (θ + pi/2)               cos θ cos pi/2  -   sin θ sin pi/2                  - sin θ  

 

 

(4/5)             - 4

____    =     ___

-(3/5)             3

 

 

I believe this is what hectictar was pointing out.......

 

cool cool cool

 May 11, 2019
edited by CPhill  May 11, 2019

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