How do you use the properties of inverse trigonometric functions to evaluate #tan(arcsin (0.31))#?

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Answer 1 · 2014-11-25T03:16:18

Let

#theta=arcsin(0.31)#

by rewriting in terms of sine,

#sin theta=0.31={O}/H =># Let #{(O=0.31),(H=1):}#

By Pythagorean Theorem,

#A=sqrt{H^2-O^2}=sqrt{1^2-(0.31)^2}=sqrt{0.9039}#

Hence,

#tan(arcsin(0.31))=tan theta={O}/A={0.31}/{sqrt{0.9039}}#

I hope that this was helpful.