Question #150d7

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Answer 1 · 2015-06-12T17:21:29

Show #cos(2arctan (1/7)) = sin (4arctan (1/3)#

#tan x = 1/7 --> x = arctan (1/7) = 8.13 -> 2x = 16.26#

#cos 2x = cos 16.26 = 0.96#

#tan y = 1/3 --> y = tan 18.43 -> 4y = 73.74# deg

#sin 4y = sin 73.74 = 0.96#

Answer 2 · 2015-06-12T17:33:24

See the proof:

We have to prove that

#cosalpha=sinbeta#, where

#alpha=2arctan(1/7)# and

#beta=4arctan(1/3)#.

So:

#tan(alpha/2)=1/7#

and

#tan(beta/4)=1/3#.

Remembering the double-angle formulae:

#sinx=(2t)/(1+t^2)#,

#cosx=(1-t^2)/(1+t^2)# where #t=tan(x/2)#

and

#sin2x=2sinxcosx#,

then:

#cosalpha=2sin(beta/2)cos(beta/2)rArr#

(where #t_1=tan(alpha/2)# and #t_2=tan(beta/4)#

#(1-t_1^2)/(1+t_1^2)=2*(2t_2)/(1+t_2^2)*(1-t_2^2)/(1+t_2^2)#

#(1-1/49)/(1+1/49)=2*(2*1/3)/(1+1/9)*(1-1/9)/(1+1/9)rArr#

#(48/49)/(50/49)=2*(2/3)/(9/10)*(8/9)/(9/10)rArr#

#48/49*49/50=2*2/3*9/10*8/9*9/10rArr#

#24/25=24/25#.