Question

Without using a graphing utility, determine the symmetries (if any) of the curve \(r=4-\sin (\theta / 2)\)

Short answer

Answer: No, the curve does not have any symmetries.

Step-by-step solution

Symmetry about the Polar Axis (\(y\)-axis)

To check if there is a symmetry about the polar axis, we have to check whether \(r\) has the same value when we replace \(\theta\) with its supplementary angle \(-\theta\). So, we evaluate the equation for \(r\), replacing \(\theta\) with \(-\theta\): \(r=4-\sin ((-\theta) / 2)\) Now, we can use the sin property that \(\sin(-x) = -\sin(x)\) and write the equation as: \(r=4+(\sin (\theta / 2))\) Comparing this with the given equation \(r=4-\sin (\theta / 2)\), we can conclude that the curve does NOT possess symmetry about the polar axis since they are not equal.

Symmetry about the Polar Axis (\(x\)-axis)

To check if there is a symmetry about the polar axis, we have to check whether \(r\) has the same value for the co-terminal angles. So, we will replace \(\theta\) with \((\theta + \pi)\) \(r=4-\sin ((\theta + \pi) / 2)\) Using sin addition formula,\(\sin(a+b) = \sin a \cos b + \cos a \sin b,\) \(r=4-\sin (\theta / 2)\cos (\pi / 2)+\cos (\theta / 2)\sin (\pi / 2)\) As we know that \(\cos (\pi / 2)=0\) and\(\sin(\pi/2) = 1\) \(r = 4 + \cos(\theta / 2)\) \(r=4\cos^2 (\theta / 4)+\sin(\theta / 2)\cos (\theta/4)-\sin(\theta/2)\cos(\theta/4)+\sin^2(\theta / 4)\) \(r=4(\cos^2 (\theta / 4)+\sin^2 (\theta / 4))\) \(r=4\) Comparing this with the given equation \(r=4-\sin (\theta / 2)\), we can conclude that the curve does NOT possess symmetry about the polar axis since they are not equal. So, the curve defined by the polar equation \(r=4-\sin (\theta / 2)\) has no symmetries.