Suppose #alpha = arctan(1/2)# and #beta = arctan(1/3)#
What is the value of #tan(alpha + beta)# ?
It's probably easiest to calculate #sin alpha#, #cos alpha#, #sin beta# and #cos beta# first:
If #alpha = arctan(1/2)#, then #1/2 = tan alpha = sin alpha / cos alpha#.
Multiplying through by #2cos alpha# we get
#cos alpha = 2 sin alpha#
Squaring both sides and using #sin^2 alpha + cos^2 alpha = 1# we get
#4 sin^2 alpha = cos^2 alpha = 1 - sin^2 alpha#
Adding #sin^2 alpha# to both sides and dividing by 5 we get
#sin^2 alpha = 1/5#
So #sin alpha = 1/sqrt(5)# and #cos alpha = 2 sin alpha = 2/sqrt(5)#.
Similarly, we can find #sin beta = 1/sqrt(10)# and #cos beta = 3/sqrt(10)#.
Now we can calculate
#tan(alpha + beta) = sin(alpha+beta)/cos(alpha+beta)#
#=(sin alpha cos beta + sin beta cos alpha)/(cos alpha cos beta - sin alpha sin beta)#
The numerator:
#sin alpha cos beta + sin beta cos alpha#
#=(1/sqrt(5))(3/sqrt(10)) + (1/sqrt(10))(2/sqrt(5))#
#=5/sqrt(50) = 1/sqrt(2)#
The denominator:
#cos alpha cos beta - sin alpha sin beta#
#(2/sqrt(5))(3/sqrt(10)) - (1/sqrt(5))(1/sqrt(10))#
#=5/sqrt(50) = 1/sqrt(2)#
So putting these together:
#tan(alpha + beta) = (1/sqrt(2))/(1/sqrt(2)) = 1#
So #alpha + beta = pi/4#
