Question 1132183
if C=arctan(3) + arcsin(5/13) find cos(C)

Without calculator
<pre>
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.

{{{drawing(350/9,100,-.2,1.6,-.4,3.2, triangle(0,0,1,0,1,3),
locate(.1,.502,A),locate(.4,0,1),locate(1.04,1.5,3) )}}}

Then we calculate the hypotenuse using the Pythagorean theorem:
{{{matrix(5,1,

c^2=a^2+b^2,
c^2=1^2+3^2,
c^2=1+9,
c^2=10,
c=sqrt(10))}}}

So the completed right triangle containing angle A is:

{{{drawing(350/9,100,-.2,1.2,-.4,3.2, triangle(0,0,1,0,1,3),
locate(.1,.502,A),locate(.4,0,1),locate(1.04,1.5,3),locate(-.2,2.07,sqrt(10)) )}}}

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.

{{{drawing(100,2900/63,-.4,12.2,-.2,5.6, triangle(0,0,12,0,12,5),
locate(3.8,1.9,B),locate(12.12,3,5),locate(3,4,13) )}}}

Then we calculate the adjacent side using the Pythagorean theorem:
{{{matrix(6,1,

c^2=a^2+b^2,
13^2=a^2+5^2,
169=a^2+25,
144=a^2,
sqrt(144)=a,
12=a)}}}

So the completed right triangle containing angle B is:

{{{drawing(100,760/13,-.4,12.2,-1.3,5.6, triangle(0,0,12,0,12,5),
locate(3.8,1.9,B),locate(12.12,3,5),locate(3,4,13),locate(6,0,12) )}}}

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 

{{{drawing(350/9,100,-.2,1.2,-.4,3.2, triangle(0,0,1,0,1,3),
locate(.1,.502,A),locate(.4,0,1),locate(1.04,1.5,3),locate(-.2,2.07,sqrt(10)) )}}}{{{drawing(100,760/13,-.4,12.2,-1.3,5.6, triangle(0,0,12,0,12,5),
locate(3.8,1.9,B),locate(12.12,3,5),locate(3,4,13),locate(6,0,12) )}}}

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

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

{{{(1/sqrt(10))(12/13)-(3/sqrt(10))(5/13)}}} =

{{{12/(13sqrt(10))-15/(13sqrt(10))}}} =

{{{-3/(13sqrt(10))}}}

If we rationalize that we get

{{{-3/(13sqrt(10))}}}{{{""*""}}}{{{sqrt(10)/sqrt(10)}}}

{{{-3sqrt(10)/130}}}

Edwin</pre>