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

1 Answer

Tangent line #y=-2x+pi#

Explanation:

Given #y=sin (2x)# at #x=pi/2#

solve for the point first

#y=sin (2x)#

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

#y=0#

Our point #(x_1, y_1)=(pi/2, 0)#

Let us solve the slope m of the tangent line

#y=sin (2x)#

#y'=cos (2x)*d/dx(2x)=cos (2x)(2)=2*cos (2x)#

#m=2*cos (2(pi/2))=2*(-1)=-2#

#m=-2#

The Tangent Line

#y-y_1=m(x-x_1)#

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

#y=-2x+pi#

Kindly see the graph of #y=sin (2x)# and #y=-2x+pi#

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God bless....I hope the explanation is useful.