What is #Cot [ arcsin( sqrt5 / 6) ]#?

1 Answer
Jul 28, 2015

#sqrt(155)/5#

Explanation:

Start by letting #arcsin(sqrt(5)/6)# to be a certain angle #alpha#

It follows that #alpha=arcsin(sqrt5/6)#
and so
#sin(alpha)=sqrt5/6#

This means that we are now looking for #cot(alpha)#

Recall that : #cot(alpha)=1/tan(alpha)=1/(sin(alpha)/cos(alpha))=cos(alpha)/sin(alpha)#

Now, use the identity #cos^2(alpha)+sin^2(alpha)=1# to obtain #cos(alpha)=sqrt((1-sin^2(alpha)))#

#=>cot(alpha)=cos(alpha)/sin(alpha)=sqrt((1-sin^2(alpha)))/sin(alpha)=sqrt((1-sin^2(alpha))/sin^2(alpha))=sqrt(1/sin^2(alpha)-1)#

Next, substitute #sin(alpha)=sqrt5/6# inside #cot(alpha)#

#=>cot(alpha)=sqrt(1/(sqrt5/6)^2-1)=sqrt(36/5-1)=sqrt(31/5)=color(blue)(sqrt(155)/5)#