How do you find the equation to the tangent line of #y=cos3x# and its tangent at the point #(pi/4, -sqrt2/2)#?

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Answer 1 · 2015-04-14T18:46:31

Have a look:
enter image source here

Answer 2 · 2015-04-14T18:49:06

Rememebering that the slope of a tangent line to a curve in a point is the derivative in that point:

#y'=-sin3x * 3#,

so

#m=y'(pi/4)=-3sin(3/4pi)=-3(sqrt2/2)#.

Than:

#y+sqrt2/2=-3sqrt2/2(x-pi/4)#.