Trigonometric ratios of angle \(cos36^{0}\)

Deriving the Exact Values of \( \sin 36^\circ \), \( \cos 36^\circ \), and \( \tan 36^\circ \)

In trigonometry, certain angles such as 36°, 18°, and 72° have exact algebraic expressions for their sine and cosine values. Let's derive the exact value of \( \sin 36^\circ \), and from there, obtain \( \cos 36^\circ \) and \( \tan 36^\circ \).

Step 1: Use a trigonometric identity

We know that: \[ \sin(3\theta) = 3\sin\theta - 4\sin^3\theta \] Let \( \theta = 36^\circ \), then \( 3\theta = 108^\circ \). Also, \( \sin(108^\circ) = \cos(18^\circ) \) since: \[ \sin(90^\circ + x) = \cos x \] So: \[ \sin(108^\circ) = \cos(18^\circ) = \frac{1}{4}\sqrt{10 + 2\sqrt{5}} \] Therefore: \[ 3\sin36^\circ - 4\sin^3 36^\circ = \frac{1}{4}\sqrt{10 + 2\sqrt{5}} \] Let \( x = \sin 36^\circ \). Then: \[ 3x - 4x^3 = \frac{1}{4}\sqrt{10 + 2\sqrt{5}} \] Solving this equation directly is complex, but we can derive \( \sin 36^\circ \) using known exact trigonometric values from geometry.

Step 2: Use a known identity from geometry

From the geometry of a regular pentagon: \[ \sin 18^\circ = \frac{\sqrt{5} - 1}{4}, \quad \sin 36^\circ = \frac{\sqrt{10 - 2\sqrt{5}}}{4} \] So, the exact value is: \[ \sin 36^\circ = \frac{\sqrt{10 - 2\sqrt{5}}}{4} \]

Step 3: Find \( \cos 36^\circ \)

Using the identity: \[ \cos^2\theta = 1 - \sin^2\theta \] We get: \[ \cos 36^\circ = \sqrt{1 - \left( \frac{\sqrt{10 - 2\sqrt{5}}}{4} \right)^2} \] Simplifying: \[ = \sqrt{1 - \frac{10 - 2\sqrt{5}}{16}} = \sqrt{ \frac{16 - (10 - 2\sqrt{5})}{16} } = \sqrt{ \frac{6 + 2\sqrt{5}}{16} } = \frac{1}{4} \sqrt{6 + 2\sqrt{5}} \] So: \[ \cos 36^\circ = \frac{1}{4} \sqrt{6 + 2\sqrt{5}} \]

Step 4: Find \( \tan 36^\circ \)

Now use: \[ \tan 36^\circ = \frac{\sin 36^\circ}{\cos 36^\circ} = \frac{ \frac{\sqrt{10 - 2\sqrt{5}}}{4} }{ \frac{1}{4} \sqrt{6 + 2\sqrt{5}} } = \frac{ \sqrt{10 - 2\sqrt{5}} }{ \sqrt{6 + 2\sqrt{5}} } \]

Summary

• \( \sin 36^\circ = \frac{\sqrt{10 - 2\sqrt{5}}}{4} \)
• \( \cos 36^\circ = \frac{1}{4} \sqrt{6 + 2\sqrt{5}} \)
• \( \tan 36^\circ = \frac{ \sqrt{10 - 2\sqrt{5}} }{ \sqrt{6 + 2\sqrt{5}} } \)