Cos(sin-¹ (-3/5)?

2 Answers
maganbhai P.
May 7, 2018

#cos(sin^-1(-3/5))=4/5#

Explanation:

We take,

#y=cos(sin^-1(-3/5))#

#=cos(-sin^-1(3/5))...color(red)(to[as,sin^-1(-x)=-sin^-1x]#

#=cos(sin^-1(3/5))...color(violet)(to[as,cos(-theta)=costheta]#

#But,color(blue)(sin^-1x=cos^-1sqrt(1-x^2)#

#:.y=cos(cos^-1(sqrt(1-(3/5)^2)))#

#=cos(cos^-1sqrt(1-9/25))#

#=cos(cos^-1 (4/5))#

#=4/5...tocolor(brown)([as,cos(cos^-1x)=x]#

#:.y=0.8#

Dean R.
May 8, 2018

#cos (arcsin (-3/5)) = \pm 4/5#

Assuming #arcsin(-3/5)# refers to any angle whose sine is #-3/5,# the sign of the cosine cannot be determined.

Explanation:

# cos (arcsin (-3/5))#

#sin^{-1}(x)#, which I prefer to denote #arcsin(x)#, refers to every inverse sine, not just the principal value, which I denote #text{Arc}text{sin}(x).#

#cos theta# where #theta = arcsin(-3/5) #

#sin theta = -3/5#

Of course #3^2 + 4^2 = 5^2 # is everybody's favorite Pythagorean Triple.

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

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

#cos theta = pm sqrt{1 - sin ^2 theta }#

#cos theta = \pm sqrt{1 - (-3/5)^2} = pm sqrt{1-9/25}= pm sqrt{16/25} = pm 4/5 #

#cos (arcsin (-3/5)) = \pm 4/5#

The sign cannot be determined with the information given.

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