Question

Determine the value of \cot(-75\degree )

Collected in the board: Trigonometry

Steven Zheng posted 5 months ago

Answer 1

First determine the value of \sin 75\degree

which is equivalent to ,

\sin (45\degree +30\degree )

=\sin 45\degree \cos 30\degree +\cos 45\degree \sin 30\degree

=\dfrac{\sqrt{2} }{2} \cdotp \dfrac{\sqrt{3} }{2}+\dfrac{\sqrt{2} }{2}\cdotp \dfrac{1}{2}

=\dfrac{\sqrt{6} +\sqrt{2} }{4}


Then,

\cos 75\degree =\sqrt{1-\sin^2 75\degree }

=\sqrt{1-(\dfrac{\sqrt{6} +\sqrt{2} }{4} )^2}

=\dfrac{\sqrt{16-8-2\sqrt{12} } }{4}

=\dfrac{\sqrt{2-\sqrt{3} } }{2}

Now \cot(-75\degree ) =-\dfrac{\cos 75\degree }{\sin 75\degree }

=-\dfrac{\dfrac{\sqrt{2-\sqrt{3} } }{2} }{\dfrac{\sqrt{6} +\sqrt{2} }{4} }

=-\dfrac{2\sqrt{2-\sqrt{3} } }{\sqrt{6}+\sqrt{2} }

=-\dfrac{\sqrt{8-4\sqrt{3} } }{\sqrt{6}+\sqrt{2} }

=-\dfrac{\sqrt{(\sqrt{6}-\sqrt{2} )^2} }{\sqrt{6}+\sqrt{2} }

=-\dfrac{\sqrt{6}-\sqrt{2} }{\sqrt{6}+\sqrt{2} }

Steven Zheng posted 5 months ago

Answer 2

Using the sum identity for cosine function to determine the value of \cos 75\degree,

which is equivalent to,

\cos(45\degree +30\degree )

=\cos 45\cos 30\degree -\sin 45\degree \sin 30\degree

=\dfrac{\sqrt{2} }{2}\cdotp \dfrac{\sqrt{3} }{2} -\dfrac{\sqrt{2} }{2}\cdotp \dfrac{1}{2}

=\dfrac{\sqrt{6}-\sqrt{2} }{4}

Now, \cot(-75\degree ) =-\dfrac{\cos 75\degree }{\sin 75\degree }

=\dfrac{\sqrt{6}-\sqrt{2} }{\sqrt{6}+\sqrt{2} }

Steven Zheng posted 5 months ago

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