How do you find the exact value of #arc tan(1/2) + arc tan (1/5) + arc tan (1/8)#?

Question

6356 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2015-06-18T14:48:27

#pi/4rad#

#arctanx+arctany+arctanz=arctan##(x+y+z-xyz)/(1-xy-yz-zx)#

Let x=#1/2#,#y=1/5#, #z=1/8#

#arctan(1/2)+arctan(1/5)+arctan(1/8)=#

#=arctan##((1/2)+(1/5)+(1/8)-(1/2*1/5*1/8))/(1-(1/2*1/5)(1/5*1/8)(1/8*1/2)#

#=arctan(0.8125/0.8125)#

#=arctan(cancel(0.8125)/cancel(0.8125))#

#=arctan(1)#

#=arctan(tan(pi/4))#[from angle table]

#=cancel(arctan)cancel(tan)((pi/4))#

#=pi/4#