Check the solution to "Solving 58cot⁡36∘=cos3x without substituting the trig values for..."

Kaspaueru2

Answered question

2022-01-16

Solving

{\frac{58}{\cot 36}}^{\circ} = \cos^{3} x

without substituting the trig values for

36^{\circ}

otoplilp1

Beginner2022-01-17Added 41 answers

Since

\cos^{3} x = \frac{3 \cos x + \cos 3 x}{4}

we need to prove that:

\frac{5 {\cos 36}^{\circ}}{8 {\sin 36}^{\circ}} = \frac{3 {\cos 18}^{\circ} + {\cos 54}^{\circ}}{4}

5 {\cos 36}^{\circ} = 3 {\sin 54}^{\circ} + 3 {\sin 18}^{\circ} + {\sin 90}^{\circ} - {\sin 18}^{\circ}

2 {\cos 36}^{\circ} + 2 {\cos 108}^{\circ} = 1

which is true because

2 {\cos 36}^{\circ} + 2 {\cos 108}^{\circ} = \frac{2 {\sin 36}^{\circ} {\cos 36}^{\circ} + 2 {\sin 36}^{\circ} {\cos 108}^{\circ}}{{\sin 36}^{\circ}} =

= \frac{{\sin 72}^{\circ} + {\sin 144}^{\circ} - {\sin 72}^{\circ}}{{\sin 36}^{\circ}} = 1

Jordan Mitchell

Beginner2022-01-18Added 31 answers

\frac{5}{8} \cdot \cot 36^{\circ} = \frac{5 \cos 36^{\circ}}{8 \sin 36^{\circ}} = \frac{5 \cos^{2} 36^{\circ}}{4 \cos 18^{\circ}}

Using Proving trigonometric equation

\cos (36^{\circ}) - \cos (72^{\circ}) = \frac{1}{2}

\cos 36^{\circ} - (2 \cos^{2} 36^{\circ} - 1) = \frac{1}{2} \Leftrightarrow 5 \cos^{2} 36^{\circ} = (1 + \cos 36^{\circ})^{2} = (2 \cos^{2} 18^{\circ})^{2}

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