How do you evaluate #tan(arcsin(1/3)) #?

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Answer 1 · 2018-04-13T01:25:37

#1/(2sqrt2)#

Recalling that #tanx=sinx/cosx,#

#tan(arcsin(1/3))=sin(arcsin(1/3))/cos(arcsin(1/3))#

#sin(arcsin(1/3))=1/3# from the fact that #sin(arcsinx)=x#, but for the cosine, some more work is needed.

Recall that

#sin^2x+cos^2x=1#

Then,

#cos^2x=1-sin^2x#

#cosx=sqrt(1-sin^2x)#

#cos(arcsin(1/3))=sqrt(1-sin^2(arcsin(1/3)))#

Knowing #sin(arcsin(1/3))=1/3, sin^2(arcsin(1/3))=(1/3)^2=1/9#

Then,

#cos(arcsin(1/3))=sqrt(1-1/9)=sqrt(8/9)=(2sqrt2)/3#

Thus,

#tan(arcsin(1/3))=(1/3)/((2sqrt2)/3)=1/cancel3*cancel3/(2sqrt2)=1/(2sqrt2)#