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

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 2 years ago

\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 2 years ago

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