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

Question

1991 views around the world

You can reuse this answer
Creative Commons License

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.