How do you find the exact value of #cos(arctan (5/2))#?

Question

↳Redirected from "How many valence electrons are in an atom of phosphorus?"

8469 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2016-01-26T09:26:20

#cos (arctan(5/2))=(2sqrt29)/29#

Let #A# be an angle whose tangent#=5/2#

Let #A=arctan(5/2)#

Then #tan A=5/2#

Imagine a right triangle with opposite side #a=5# and adjacent side #b=2#. Compute hypotenuse #c#

#c=sqrt(a^2+b^2)#

#c=sqrt(5^2+2^2)#

#c=sqrt29#

Then, the cosine function of #A#

#cos A=b/c=2/sqrt29=(2sqrt29)/29#

Have a nice day !!! From the Philippines...