Question

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

Answer 1

Simplify before you differentiate.

Explanation:

By the Cosine Sum Formula
`cos(theta + pi/2) = cos(theta)cos(pi/2)-sin(theta)sin(pi/2) = 0 -sin(theta) = -sin(theta)`
By the Cosine Difference Formula
`cos(theta - pi) = cos(theta)cos(pi)+sin(theta)sin(pi) = -cos(theta)+0 = -cos(theta)`
Their product is
`r = (-sin(theta))(-cos(theta))=sin(theta)cos(theta)`
By the Sine Double angle formula, we have
`r = sin(theta)cos(theta) = (1/2)sin(2theta)`

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 = -pi/8`. Hint: `x = rcos(theta) and y = rsin(theta)`. Once you have a point and the slope, use the Point-Slope Form of a line to obtain its equation.