How do you find the equation of the line tangent to #y=sin(2x)# at x=pi/2?

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Answer 1 · 2016-01-15T11:29:22

#y=-2x+ pi#

#y=sin (2x)#

#y=sin (2*pi/2)#
#y=0#

the point #(pi/2, 0)#

#y' = 2* cos (2x)#
#y'=# slope
#y' =2* cos 2(pi/2)#

slope =#-2#

#y-0=-2(x-pi/2)#

#y=-2x+ pi#