What is the tangent of y=sin(pi-x) ?

The tangent of y=sin(pi-x) is y=cos(a_{0})x+sin(pi-a_{0})-cos(a_{0})a_{0}

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tangent of y=sin(pi-x)

tangent of

y = sin (π - x)

y = cos (a_{0}) x + sin (π - a_{0}) - cos (a_{0}) a_{0}

Solution steps

Compute the tangent line to the general point

x = a_{0}

Find the tangent point:

(a_{0}, sin (π - a_{0}))

Find the slope of

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

Compute the slope of

y = sin (π - x)

x = a_{0} : m = cos (a_{0})

Find the line with slope m=

cos (a_{0})

and passing through

(a_{0}, sin (π - a_{0})) : y = cos (a_{0}) x + sin (π - a_{0}) - cos (a_{0}) a_{0}

y = cos (a_{0}) x + sin (π - a_{0}) - cos (a_{0}) a_{0}

f (x) = 7 x - 17, at x = 3

\frac{d}{d x} (6 x^{\frac{1}{3}} + 3 x^{\frac{4}{3}})

\int \frac{x + 1}{x} d x

(x^{2} - 2 x + 3)^{^{'}}

\int_{- 1}^{0} x^{3} d x

What is the tangent of y=sin(pi-x) ?
The tangent of y=sin(pi-x) is y=cos(a_{0})x+sin(pi-a_{0})-cos(a_{0})a_{0}