+0  
 
0
321
3
avatar

How do you evaluate cos(tan^-1 (1/4) + cos^-1 (1/2))?

Guest Mar 19, 2017
 #2
avatar+87334 
+1

arctan (1/4) =  14.04°

 

arccos (1/2)   = 60°

 

So

 

cos ( arctan (1/4)  + arccos (1/2) )   =

 

cos (14.04 + 60)  =

 

cos (74.04)   ≈ 0.275

 

 

cool cool cool

CPhill  Mar 20, 2017
 #3
avatar
+1

Find the third side of the right-angled triangles and read off the other two ratios,

Angle A = \(\tan^{-1}(1/4)=\cos^{-1}(4/\sqrt{17})=\sin^{-1}(1/\sqrt{17}),\)

Angle B = \(\cos^{-1}(1/2)=\sin^{-1}(\sqrt{3}/2)=\tan^{-1}\sqrt{3}\).

\(\displaystyle \cos(A+B)= \cos(A)\cos(B)-\sin(A)\sin(B)\\=\frac{4}{\sqrt{17}}.\frac{1}{2}-\frac{1}{\sqrt{17}}.\frac{\sqrt{3}}{2}=\frac{4-\sqrt{3}}{2\sqrt{17}}\approx0.275029.\)

Guest Mar 20, 2017

5 Online Users

avatar

New Privacy Policy

We use cookies to personalise content and advertisements and to analyse access to our website. Furthermore, our partners for online advertising receive information about your use of our website.
For more information: our cookie policy and privacy policy.