What is #cos[Arcsin(-4/5)]#?

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Answer 1 · 2015-07-21T22:40:38

#3/5#

First consider that : #epsilon=arcsin(-4/5)#

#epsilon# simply represents an angle.

This means that we are looking for #color(red)cos(epsilon)!#

If #epsilon=arcsin(-4/5)# then,

#=>sin(epsilon)=-4/5#

To find #cos(epsilon)# We use the identity : #cos^2(epsilon)=1-sin^2(epsilon)#

#=>cos(epsilon)=sqrt(1-sin^2(epsilon)#

#=>cos(epsilon)=sqrt(1-(-4/5)^2)=sqrt((25-16)/25)=sqrt(9/25)=color(blue)(3/5)#