What is the equation of the tangent line for #sin(x)# at the point x=2?

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Answer 1 · 2015-04-22T01:00:42

#y=xcos(2) - 2cos(2) + sin(2)#

or, if you prefer:
#y=xcos(2) + (sin(2) - 2cos(2))#

For #f(x)=sinx#, we know #f'(x)=cosx#,

so at the point #(2, sin(2))#, the slope of the tangent line is:

#m=f'(2) = cos(2)#

The equation of the line through #(2, sin(2))#, with slope #cos(2)# is

#y=xcos(2) - 2cos(2) + sin(2)#