Find exact value of the expression? sin[tan^-1(2)]

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Answer 1 · 2018-07-25T06:21:27

#sin(tan^(-1)(2))=2/sqrt5#

Let #alpha=tan^(-1)2#, then #tanalpha=2#

and #cotalpha=1/2# and #cscalpha=sqrt(1+(1/2)^2)=sqrt5/2#

Hence #sin(tan^(-1)(2))=sinalpha=2/sqrt5#

Answer 2 · 2018-07-25T06:21:40

#sin[tan^-1(2)]=2/sqrt5#

We know that,

#color(red)((1)tan^-1x=arc tanx=arc sin(x/sqrt(1+x^2)) # , #x > 0#

#color(blue)((2)sin(sin^-1x)=sin(arc sinx)=x # , #x in [-1,1]#

Let ,

#A=sin[tan^-1(2)]=sin(arc tan2)#

#:.A=sin(color(red)(arc sin (2/sqrt(1+2^2))))...to color(red)(Aplly (1)#

#:.A=color(blue)(sin(arc sin(2/sqrt5))).....tocolor(blue)(Apply(2)#

#:.A=2/sqrt5 ,where ,2/sqrt5~~0.8944 in [-1,1]#

Answer 3 · 2018-07-25T06:25:36

#2/sqrt5#.

Both #( sin^(-1)) and (tan^(-1))# values #in ( - pi/2, pi/2 )#.

If tan value #> 0#, so is sine. And so, .

#tan^(-1)( 2 ) = sin^(-1)( 2/sqrt5)#. Now,

#sin tan^(-1)(2 ) = sin sin^(-1)(2/sqrt5) = (1)(2/ sqrt5)#,

using

# sin sin^(-1), cos cos^(-1), tan tan^(-1),#

#sec sec^(-1), csc csc^(-1) and cot cot^(-1)#

operations amount to multiplication by 1.