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

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

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

tangent of

y = sin (sin (x)), at x = π

y = - x + π

Solution steps

Find the tangent point:

(π, 0)

Find the slope of

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

Compute the slope of

y = sin (sin (x))

x = π : m = - 1

Find the line with slope m=

- 1

and passing through

(π, 0) : y = - x + π

y = - x + π

y = x, (9, 3)

\frac{sin ( x ) + cos ( x )}{cos ( x )}

y = \frac{8 x}{x ^{2} + 1}, (- 1, - 4)

\int \frac{1}{x ^{2} + 2 x - 24} d x

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