For #f(t)= (sin^2t,cos^2t)# what is the distance between #f(pi/4)# and #f(pi)#?

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Answer 1 · 2017-03-16T13:52:01

The answer is #=sqrt2/2=0.707#

#cos(pi/4)=sqrt2/2#

#sin(pi/4)=sqrt2/2#

#sinpi=0#

#cospi=-1#

Let's calculate #f(pi/4)# and #f(pi)#

#f(t)=(sin^2t, cos^2t)#

#f(pi/4)=(sin^2(pi/4), cos^2(pi/4))#

#=((sqrt2/2)^2,(sqrt2/2)^2)#

#=(1/2,1/2)#

#f(pi)=(sin^2(pi), cos^2(pi))#

#=(0,1)#

The distance between 2 points #(x_1,y_1)# and #(x_2,y_2)# is

#d=sqrt((x_2-x_1)^2+y_2-y_1)^2)#

The distance is

#=sqrt((0-1/2)^2+(1-1/2)^2)#

#=sqrt(1/4+1/4)#

#=sqrt2/2=0.707#