How do you find the equation of the line tangent to #y=sinx# at #(pi/4, sqrt(2)/2)#?

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Answer 1 · 2016-09-24T20:14:46

#y=(4x-pi+4)/(4sqrt2)#

The slope of the tangent line can be found using the derivative of the function:

#dy/dx=cosx#

The slope of the tangent line at #x=pi/4# can then be found through plugging #pi/4# into the derivative.

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

So, we have a line that passes through the point #(pi/4,sqrt2/2)# and a slope of #sqrt2/2#:

#y-y_0=m(x-x_0)#

#y-sqrt2/2=sqrt2/2(x-pi/4)#

#y=sqrt2/2x-(pisqrt2)/8+sqrt2/2#

#y=(4sqrt2x-pisqrt2+4sqrt2)/8#

Note that #sqrt2/8=sqrt2/(4sqrt2sqrt2)=1/(4sqrt2)#:

#y=(4x-pi+4)/(4sqrt2)#