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

Simplify before you differentiate.

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

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To find the equation of the tangent line to the polar curve ( r = \cos\left(\theta + \frac{\pi}{2}\right) \cdot \cos\left(\theta - \pi\right) ) at ( \theta = -\frac{\pi}{8} ), we first find the derivative of ( r ) with respect to ( \theta ), denoted as ( \frac{dr}{d\theta} ). Then we evaluate ( r ) and ( \frac{dr}{d\theta} ) at ( \theta = -\frac{\pi}{8} ) to get the slope of the tangent line.

[ r = \cos\left(\theta + \frac{\pi}{2}\right) \cdot \cos\left(\theta - \pi\right) ]

[ \frac{dr}{d\theta} = \frac{d}{d\theta}\left[\cos\left(\theta + \frac{\pi}{2}\right) \cdot \cos\left(\theta - \pi\right)\right] ]

[ \frac{dr}{d\theta} = -\sin\left(\theta + \frac{\pi}{2}\right) \cdot \cos\left(\theta - \pi\right) - \cos\left(\theta + \frac{\pi}{2}\right) \cdot (-\sin\left(\theta - \pi\right)) ]

[ \frac{dr}{d\theta} = -\sin\left(\theta + \frac{\pi}{2}\right) \cdot \cos\left(\theta - \pi\right) + \cos\left(\theta + \frac{\pi}{2}\right) \cdot \sin\left(\theta - \pi\right) ]

[ \frac{dr}{d\theta} = -\sin\left(\theta + \frac{\pi}{2}\right) \cdot \cos\left(\theta - \pi\right) + \cos\left(\theta + \frac{\pi}{2}\right) \cdot \sin\left(\theta - \pi\right) ]

[ \frac{dr}{d\theta} = -\sin\left(\theta + \frac{\pi}{2}\right) \cdot \cos\left(\theta - \pi\right) + \cos\left(\theta + \frac{\pi}{2}\right) \cdot \sin\left(\theta - \pi\right) ]

[ \frac{dr}{d\theta} = -\cos\theta \cdot \cos\left(\theta - \pi\right) + (-\sin\theta) \cdot \sin\left(\theta - \pi\right) ]

[ \frac{dr}{d\theta} = -\cos\theta \cdot \cos\theta + (-\sin\theta) \cdot (-\sin\theta) ]

[ \frac{dr}{d\theta} = -\cos^2\theta - \sin^2\theta ]

[ \frac{dr}{d\theta} = -1 ]

The slope of the tangent line at ( \theta = -\frac{\pi}{8} ) is ( -1 ). Using the point-slope form of the equation of a line, we have:

[ y - y_1 = m(x - x_1) ]

[ y - r(-\frac{\pi}{8}) = -1(\theta - (-\frac{\pi}{8})) ]

[ y + \frac{\sqrt{2}}{2} = -(\theta + \frac{\pi}{8}) ]

[ y = -\theta - \frac{\pi}{8} - \frac{\sqrt{2}}{2} ]

So, the equation of the tangent line to the polar curve ( r = \cos\left(\theta + \frac{\pi}{2}\right) \cdot \cos\left(\theta - \pi\right) ) at ( \theta = -\frac{\pi}{8} ) is ( y = -\theta - \frac{\pi}{8} - \frac{\sqrt{2}}{2} ).

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