C = arctan(3) + arcsin(5/13) find cos(C)

Let A = arctan(3) which means tan(A) = 3

So we draw a right triangle that has angle A.  Since the tangent is
opposite/adjacent and since tan(A) = 3/1 we will make the opposite
side 3 and the adjacent side 1 so that tan(A) = 3/1.



Then we calculate the hypotenuse using the Pythagorean theorem:


So the completed right triangle containing angle A is:



Let B = arcsin(5/13) which means sin(B) = 5/13

So we draw a right triangle that has angle B.  Since the sine is
opposite/hypotenuse and since sin(B) = 5/13 we will make the opposite
side 5 and the hypotenuse 13 so that sin(B) = 5/13.



Then we calculate the adjacent side using the Pythagorean theorem:


So the completed right triangle containing angle B is:



Since C = arctan(3) + arcsin(5/13), and since we let

A = arctan(3) which means tan(A) = 3, and
B = arcsin(5/13) which means sin(B) = 5/13

Then C = A + B

We want cos(C) which is cos(A + B), we use the identity

cos(A + B) = cos(A)cos(B)-sin(A)sin(A)

We use the two right triangles 



and the fact that
cosine = adjacent/hypotenuse and sine = opposite/hypotenuse

cos(C) = cos(A + B) = cos(A)cos(B)-sin(A)sin(A) =

 =

 =



If we rationalize that we get





Edwin