What is the tangent of r=cos(2θ) ?

The tangent of r=cos(2θ) is r=-sin(2a_{0})*2θ+cos(2a_{0})+2a_{0}sin(2a_{0})

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tangent of r=cos(2θ)

tangent of

r = cos (2 θ)

r = - sin (2 a_{0}) \cdot 2 θ + cos (2 a_{0}) + 2 a_{0} sin (2 a_{0})

Solution steps

Compute the tangent line to the general point

θ = a_{0}

Find the tangent point:

(a_{0}, cos (2 a_{0}))

Find the slope of

r = cos (2 θ) : \frac{d r}{d θ} = - sin (2 θ) \cdot 2

Compute the slope of

r = cos (2 θ)

θ = a_{0} : m = - sin (2 a_{0}) \cdot 2

Find the line with slope m=

- sin (2 a_{0}) 2

and passing through

(a_{0}, cos (2 a_{0})) : r = - sin (2 a_{0}) \cdot 2 θ + cos (2 a_{0}) + 2 a_{0} sin (2 a_{0})

r = - sin (2 a_{0}) \cdot 2 θ + cos (2 a_{0}) + 2 a_{0} sin (2 a_{0})

\frac{\partial}{\partial y} (y^{4} sin (3 x))

\int \frac{1}{x ^{2} + 2 x + 6} d x

\int (e^{2 t} + \frac{1}{( t + 1 ) ^{2}}) d t

\frac{s - 3}{s ^{2} + 10 s + 9}

x \to 4 lim (2 x + 1)

What is the tangent of r=cos(2θ) ?
The tangent of r=cos(2θ) is r=-sin(2a_{0})*2θ+cos(2a_{0})+2a_{0}sin(2a_{0})