inverse trig (1 Viewer)

[jemsta]

hey guys
jus need one help on a question
prove:
2 tan^-1 2= pi - cos^-1 3/5

 

[yrtherenonames]

hmm it's quite doable lemme draw you a diagram one sec...ahh actually you can draw it yourself.
HOK
Let A=tan^-1(2) Let B=cos^-1(3/5)
tanA=2 cosB=3/5
Rearranging,
LHS=2A+B RHS=pi
sin(RHS)=sin(pi)=0
Required to prove that sin(LHS)=sin(2A+B)=0
sin(2A+B)=sin2AcosB+cos2AsinB
here you need a diagram, but after that it comes out as
(2)(2/rt5)(1/rt5)(3/5)+.8(-3/5)
=0
Sorry I'm no good with diagrams but if you draw it yourself you'll see it comes out.
The dude below me has a pretty solid method too.

 

[香港!]

hey guys
jus need one help on a question
prove:
2 tan^-1 2= pi - cos^-1 3/5

tan (pi-cos^-1 (3\5))=[tan pi-tan (cos^-1 3\5)]\(1+tan pi tan(cos^-1 3\5) )

let cos^-1 3\5=A
cosA=3\5
hence: tanA=4\3

tan (pi-cos^-1 (3\5))=-tan A\(1+0)=-tanA=-4\3

let tan^-1 2=B
tan B=2
tan (2B)=2tanB\(1-tan²B)
=2(2)\(1-4)=4\-3=-4\3
.: 2 tan^-1 2= pi - cos^-1 3/5