Question

Determine the value of \sin 6\degree \cdotp \sin 42\degree\cdotp \sin 66\degree\cdotp \sin 78\degree

Collected in the board: Trigonometry

Steven Zheng posted 4 months ago

Answer

\sin 6\degree \cdotp \sin 42\degree\cdotp \sin 66\degree\cdotp \sin 78\degree

= \sin 6\degree \cdotp \sin(90\degree-48\degree)\cdotp \sin(90\degree- 24\degree)\cdotp \sin(90\degree- 12\degree)

= \sin 6\degree\cos 12\degree \cos 24\degree \cos 48\degree

=\dfrac{ 2\sin 6\degree\cos 6\degree \cos 12\degree \cos 24\degree \cos 48\degree }{2\cos 6\degree }

=\dfrac{\sin 12\degree \cos 12\degree \cos 24\degree \cos 48\degree }{2\cos 6\degree}

=\dfrac{\sin 24\cos 24\degree \cos 48\degree }{4\cos 6\degree}

=\dfrac{\sin 48\degree \cos 48\degree }{8\cos 6\degree}

=\dfrac{\sin 96\degree }{16\cos 6\degree}

=\dfrac{\cos 6\degree }{16\cos 6\degree}

=\dfrac{1}{16}

Steven Zheng posted 4 months ago

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