What is the equation of the tangent line of r=cos(theta+pi/2)*cos(theta-pi)  at theta=(-pi)/8?

Mar 25, 2017

Simplify before you differentiate.

Explanation:

By the Cosine Sum Formula
$\cos \left(\theta + \frac{\pi}{2}\right) = \cos \left(\theta\right) \cos \left(\frac{\pi}{2}\right) - \sin \left(\theta\right) \sin \left(\frac{\pi}{2}\right) = 0 - \sin \left(\theta\right) = - \sin \left(\theta\right)$
By the Cosine Difference Formula
$\cos \left(\theta - \pi\right) = \cos \left(\theta\right) \cos \left(\pi\right) + \sin \left(\theta\right) \sin \left(\pi\right) = - \cos \left(\theta\right) + 0 = - \cos \left(\theta\right)$
Their product is
$r = \left(- \sin \left(\theta\right)\right) \left(- \cos \left(\theta\right)\right) = \sin \left(\theta\right) \cos \left(\theta\right)$
By the Sine Double angle formula, we have
$r = \sin \left(\theta\right) \cos \left(\theta\right) = \left(\frac{1}{2}\right) \sin \left(2 \theta\right)$

To find the slope of the tangent line, use the fact that

dy/dx=((dr)/(d(theta))sin(theta)+rcos(theta))/((dr)/(d(theta))cos(theta)-rsin(theta)

Finally, determine the point (x, y) corresponding to the point that you are interested in, where $\theta = - \frac{\pi}{8}$. Hint: $x = r \cos \left(\theta\right) \mathmr{and} y = r \sin \left(\theta\right)$. Once you have a point and the slope, use the Point-Slope Form of a line to obtain its equation.