How do you evaluate # cos(arctan(2) +arctan(3))#?

2 Answers
Apr 16, 2016

#cos(arctan(2)+arctan(3)) = -sqrt(2)/2#

Explanation:

Consider the following two right angled triangles:

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Then:

#{ (tan(alpha) = 2), (tan(beta) = 3) :}#

Notice that:

#{ (sin(alpha) = 2/sqrt(5)), (cos(alpha) = 1/sqrt(5)), (sin(beta) = 3/sqrt(10)), (cos(beta) = 1/sqrt(10)) :}#

The sum of angles formula for #cos# tells us:

#cos(alpha+beta) = cos(alpha) cos(beta) - sin(alpha) sin(beta)#

#=1/sqrt(5) 1/sqrt(10) - 2/sqrt(5) 3/sqrt(10)#

#=(1-6)/sqrt(50)=-5/(5sqrt(2))=-1/sqrt(2)=-sqrt(2)/2#

Jul 16, 2016

#cos(tan^-1 2+tan^-1 3)#

#=cos(tan^-1( (2+ 3)/(1-2*3)))#

#=cos(tan^-1( -1))#

#=cos(tan^-1( -tan(pi/4)))#

#=cos(tan^-1( tan(pi-pi/4)))#

#=cos(pi-pi/4)#

#=-cos(pi/4)#

#=-1/sqrt2#