What is the tangent of 3cos(3θ) ?

The tangent of 3cos(3θ) is y=-9sin(3a_{0})θ+3cos(3a_{0})+9a_{0}sin(3a_{0})

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tangent of 3cos(3θ)

tangent of

3 cos (3 θ)

y = - 9 sin (3 a_{0}) θ + 3 cos (3 a_{0}) + 9 a_{0} sin (3 a_{0})

Solution steps

Compute the tangent line to the general point

θ = a_{0}

Find the tangent point:

(a_{0}, 3 cos (3 a_{0}))

Find the slope of

y = 3 cos (3 θ) : \frac{d y}{d θ} = - 9 sin (3 θ)

Compute the slope of

y = 3 cos (3 θ)

θ = a_{0} : m = - 9 sin (3 a_{0})

Find the line with slope m=

- 9 sin (3 a_{0})

and passing through

(a_{0}, 3 cos (3 a_{0})) : y = - 9 sin (3 a_{0}) θ + 3 cos (3 a_{0}) + 9 a_{0} sin (3 a_{0})

y = - 9 sin (3 a_{0}) θ + 3 cos (3 a_{0}) + 9 a_{0} sin (3 a_{0})

\int_{0}^{5} v (t) d t

\frac{\partial}{\partial x} ((x y + 2) e^{- x y})

\frac{d}{d x} (- 4 sin (x) cos (x))

\int \frac{3}{x + 4} d x

f (x) = x^{3} + x, at x = 2

What is the tangent of 3cos(3θ) ?
The tangent of 3cos(3θ) is y=-9sin(3a_{0})θ+3cos(3a_{0})+9a_{0}sin(3a_{0})