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,

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

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