Respuesta :
now, recall that, testing for symmetry is done by replacing the argument, and if the resulting function mimics the original function, then the function has symmetry of that type.
so let's test r = -1 - 3sin(θ).
[tex] \bf recall
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\textit{Sum and Difference Identities}
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sin(\alpha + \beta)=sin(\alpha)cos(\beta) + cos(\alpha)sin(\beta)
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sin(\alpha - \beta)=sin(\alpha)cos(\beta)- cos(\alpha)sin(\beta)
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and\qquad \qquad \stackrel{\textit{Symmetry Identities}}{sin(-\theta )=-sin(\theta)} [/tex]
also recall that sin(π) = 0, and cos(π) = -1.
with that in mind, let's check
[tex] \bf \stackrel{\textit{testing for the y-axis, }\theta =\pi -\theta }{r=-1-3sin(\pi -\theta )}
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r=-1-3[sin(\pi )cos(\theta )-cos(\pi )sin(\theta )]
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r=-1-3[0cos(\theta )-(-1)sin(\theta )]\implies r=-1-3[0+sin(\theta )]
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r=-1-3sin(\theta )\quad \checkmark\\\\
------------------------------- [/tex]
[tex] \bf \stackrel{\textit{testing for the x-axis, }\theta =-\theta }{r=-1-3sin(-\theta )}\implies r=-1-3[-sin(\theta )]
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r=-1+3sin(\theta )\quad \otimes
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\stackrel{\textit{testing for the origin, }\theta =\pi +\theta }{r=-1-3sin(\pi +\theta )}
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r=-1-3[sin(\pi )cos(\theta )+cos(\pi )sin(\theta )]
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r=-1-3[0cos(\theta )+(-1)sin(\theta )]\implies r=-1-3[0-sin(\theta )]
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r=-1+3sin(\theta )\quad \otimes [/tex]
