Trigonometry! #39

Geometry Level 3

Without using tables, determine the value of csc ⁡ 1 2 ∘ csc ⁡ 4 8 ∘ csc ⁡ 5 4 ∘ \csc 12^{\circ} \csc 48^{\circ} \csc 54^{\circ}

This problem is part of the set Trigonometry .


The answer is 8.

Omkar Kulkarni by Omkar Kulkarni , 22, India

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2 solutions

Mas Mus
Mar 12, 2015

Let A = sin ⁡ 1 2 ∘ sin ⁡ 4 8 ∘ sin ⁡ 5 4 ∘ A=\sin 12^{\circ} \sin 48^{\circ} \sin 54^{\circ}

= − 1 2 [ cos ⁡ 6 0 ∘ − cos ⁡ 3 6 ∘ ] sin ⁡ 5 4 ∘ =-\frac{1}{2}[\cos 60^{\circ}- \cos 36^{\circ}] \sin 54^{\circ}

= − 1 4 sin ⁡ 5 4 ∘ + 1 2 cos ⁡ 3 6 ∘ sin ⁡ 5 4 ∘ =-\frac{1}{4}\sin 54^{\circ}+\frac{1}{2}\cos 36^{\circ} \sin 54^{\circ}

= − 1 4 sin ⁡ 5 4 ∘ + 1 4 [ sin ⁡ 9 0 ∘ + sin ⁡ 1 8 ∘ ] =-\frac{1}{4}\sin 54^{\circ}+\frac{1}{4}[\sin 90^{\circ} +\sin 18^{\circ}]

= 1 4 − 1 4 [ sin ⁡ 5 4 ∘ − sin ⁡ 1 8 ∘ ] = 1 4 − 1 4 [ cos ⁡ 3 6 ∘ sin ⁡ 3 6 ∘ cos ⁡ 1 8 ∘ ] =\frac{1}{4}-\frac{1}{4}[\sin54^{\circ}-\sin18^{\circ}]=\frac{1}{4}-\frac{1}{4}[\frac{\cos36^{\circ}\sin36^{\circ}}{\cos18^{\circ}}]

= 1 4 − 1 4 × 1 2 [ sin ⁡ 7 2 ∘ cos ⁡ 1 8 ∘ ] = 1 4 − 1 8 = 1 8 =\frac{1}{4}-\frac{1}{4}\times\frac{1}{2}[\frac{\sin72^{\circ}}{\cos18^{\circ}}]=\frac{1}{4}-\frac{1}{8}=\frac{1}{8}

Now, csc ⁡ 1 2 ∘ csc ⁡ 4 8 ∘ csc ⁡ 5 4 ∘ = 1 A = 8 \csc 12^{\circ} \csc 48^{\circ} \csc 54^{\circ}=\frac{1}{A}=\boxed8

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Omkar Kulkarni
Feb 14, 2015

sin ⁡ 1 2 ∘ sin ⁡ 4 8 ∘ sin ⁡ 5 4 ∘ = 1 2 ( 2 sin ⁡ 1 2 ∘ sin ⁡ 4 8 ∘ ) sin ⁡ 5 4 ∘ = − 1 2 ( cos ⁡ 3 6 ∘ − cos ⁡ 6 0 ∘ ) sin ⁡ 5 4 ∘ = − 1 2 ( cos ⁡ 3 6 ∘ − 1 2 ) sin ⁡ 5 4 ∘ = − 1 4 ( 2 cos ⁡ 3 6 ∘ sin ⁡ 5 4 ∘ − sin ⁡ 5 4 ∘ ) = − 1 4 ( sin ⁡ 9 0 ∘ + sin ⁡ 1 8 ∘ − sin ⁡ 5 4 ∘ ) = − 1 4 ( 1 + 5 − 1 4 − 5 + 1 4 ) = − 1 4 ( 1 + 5 − 1 − 5 − 1 4 ) = − 1 4 ( 1 − 1 2 ) = 1 8 \sin12^{\circ}\sin48^{\circ}\sin54^{\circ} \\ =\frac{1}{2}(2\sin12^{\circ}\sin48^{\circ})\sin54^{\circ} \\ = -\frac{1}{2}(\cos36^{\circ}-\cos60^{\circ})\sin54^{\circ} \\ = -\frac{1}{2}\left(\cos36^{\circ}-\frac{1}{2}\right)\sin54^{\circ} \\ = -\frac{1}{4}(2\cos36^{\circ}\sin54^{\circ}-\sin54^{\circ}) \\ = -\frac{1}{4}(\sin90^{\circ}+\sin18^{\circ}-\sin54^{\circ}) \\ = -\frac{1}{4}\left(1+\frac{\sqrt{5}-1}{4}-\frac{\sqrt{5}+1}{4}\right) \\ = -\frac{1}{4}\left(1+\frac{\sqrt{5}-1-\sqrt{5}-1}{4}\right) \\ = -\frac{1}{4}\left(1-\frac{1}{2}\right) \\ = \boxed{\frac{1}{8}}

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