What is the tangent of f(x)=sin(x),\at x=pi ?

The tangent of f(x)=sin(x),\at x=pi is y=-x+pi

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tangent of f(x)=sin(x),\at x=pi

tangent of

f (x) = sin (x), at x = π

y = - x + π

Solution steps

Find the tangent point:

(π, 0)

Find the slope of

f (x) = sin (x) : \frac{df ( x )}{d x} = cos (x)

Compute the slope of

f (x) = sin (x)

x = π : m = - 1

Find the line with slope m=

- 1

and passing through

(π, 0) : y = - x + π

y = - x + π

e^{x} sin (x), 0

\frac{d}{d x} (e^{e} + 2 ln (1 + x^{2}) - 3^{x})

\int \frac{3 x}{6 x ^{2} + 9} d x

f (x) = - \frac{2 x - 10}{x - 3}, at x = 2

y = x^{2}, x = y^{2}, y = 1

What is the tangent of f(x)=sin(x),\at x=pi ?
The tangent of f(x)=sin(x),\at x=pi is y=-x+pi