What is the tangent of f(x)=sin(x),\at x=(7pi)/6 ?

The tangent of f(x)=sin(x),\at x=(7pi)/6 is y=-(sqrt(3))/2 x-1/2+(7pi)/(4sqrt(3))

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tangent of f(x)=sin(x),\at x=(7pi)/6

tangent of

f (x) = sin (x), at x = \frac{7 π}{6}

y = - \frac{3}{2} x - \frac{1}{2} + \frac{7 π}{4 3}

Solution steps

Find the tangent point:

(\frac{7 π}{6}, - \frac{1}{2})

Find the slope of

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

Compute the slope of

f (x) = sin (x)

x = \frac{7 π}{6} : m = - \frac{3}{2}

Find the line with slope m=

- \frac{3}{2}

and passing through

(\frac{7 π}{6}, - \frac{1}{2}) : y = - \frac{3}{2} x - \frac{1}{2} + \frac{7 π}{4 3}

y = - \frac{3}{2} x - \frac{1}{2} + \frac{7 π}{4 3}

x \to 2 lim (\frac{8 - x ^{3}}{x ^{2} - 2 x})

\int \frac{x + x}{x ^{2}} d x

\frac{d}{d x} (7 sin (x) - 7 x cos (x))

\int_{0}^{π} 2 sin (x) d x

What is the tangent of f(x)=sin(x),\at x=(7pi)/6 ?
The tangent of f(x)=sin(x),\at x=(7pi)/6 is y=-(sqrt(3))/2 x-1/2+(7pi)/(4sqrt(3))