Question

Verify that \tan 3\theta = 4\tan \theta\cdotp \tan(60\degree -\theta )\cdotp \tan(60\degree +\theta )

Collected in the board: Trigonometry

Steven Zheng Steven Zheng posted 2 weeks ago


Answer

4\tan \theta\cdotp \tan(60\degree -\theta )\cdotp \tan(60\degree +\theta )

=4\cdotp \dfrac{\sin \theta\cdotp \sin(60\degree -\theta )\cdotp \sin(60\degree +\theta ) }{\cos \theta\cdotp \cos(60\degree -\theta )\cdotp \cos(60\degree +\theta ) }

4\sin \theta \cdotp \sin(60\degree -\theta )\cdotp \sin(60+\theta )

=2\sin \theta (\cos2\theta -\cos 120\degree )

=2\sin \theta (\cos2\theta+\dfrac{1}{2} )

=2\sin \theta (\dfrac{3}{2}-2\sin^2 \theta )

=3\sin \theta -4\sin^3 \theta

=\sin 3\theta

4\cos \theta \cdotp \cos(60\degree -\theta )\cos(60\degree +\theta )

=2\cos \theta (\cos 120\degree +\cos 2\theta )

=2\cos \theta (-\dfrac{1}{2}+2\cos^2 \theta -1)

=-3\cos \theta +4\cos^3 \theta

=\cos 3\theta

Therefore,

4\tan \theta\cdotp \tan(60\degree -\theta )\cdotp \tan(60\degree +\theta ) =\tan 3\theta

Steven Zheng Steven Zheng posted 2 weeks ago

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