Resolved: "Show that :31+815=16(1+cos⁡6∘)(1+cos⁡42∘)(1+cos⁡66∘)(1−cos⁡78∘)"

demonlikw4

Answered question

2022-02-27

Show that :

31 + 8 \sqrt{15} = 16 (1 + {\cos 6}^{\circ}) (1 + {\cos 42}^{\circ}) (1 + {\cos 66}^{\circ}) (1 - {\cos 78}^{\circ})

Asa Buck

Beginner2022-02-28Added 8 answers

(1 + {\cos 6}^{\circ}) (1 + {\cos 42}^{\circ}) (1 + {\cos 66}^{\circ}) (1 - {\cos 78}^{\circ})

= (1 + {\cos 6}^{\circ}) (1 + {\cos 66}^{\circ}) (1 - {\cos 78}^{\circ}) (1 + {\cos 42}^{\circ})

= (1 + {\cos 6}^{\circ} + {\cos 66}^{\circ} + {\cos 6}^{\circ} {\cos 66}^{\circ}) (1 + {\cos 42}^{\circ} - {\cos 78}^{\circ} - {\cos 78}^{\circ})

= [1 + 2 {\cos 30}^{\circ} {\cos 36}^{\circ} + \frac{1}{2} ({\cos 60}^{\circ} + {\cos 72}^{\circ})] [1 + 2 {\sin 18}^{\circ} {\sin 60}^{\circ} - \frac{1}{2} ({\cos 36}^{\circ} + {\cos 120}^{\circ})]

(Applying

2 \cos A \cos B = \cos (A - B) + \cos (A + B), \cos 2 C + \cos 2 D = 2 \cos (C - D) \cos (C + D)

and

\cos 2 C - \cos 2 D = - 2 \sin (C - D) \sin (C + D)

)
Now

{\sin 60}^{\circ} = {\cos 30}^{\circ} = \frac{\sqrt{3}}{2}, {\cos 120}^{\circ} = {\cos (180 - 60)}^{\circ} = - {\cos 60}^{\circ} = - \frac{1}{2}

So,

{\cos 72}^{\circ} = {\sin 18}^{\circ} = \frac{\sqrt{5} - 1}{4}

and

{\cos 36}^{\circ} = \frac{\sqrt{5} + 1}{4}

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