Finding the Value of sin 36°

To find the value of sin 36°, we can use trigonometric identities by setting up an equation involving the angle 36° and its multiples. A common approach is to consider the relationship between 36°, 72° (which is 2 × 36°), and 108° (which is 3 × 36°).

Derivation Steps for sin 36°

Let \(\theta = 36^\circ\). Then \(5\theta = 180^\circ\). We can split this into \(2\theta = 180^\circ - 3\theta\).

Taking the sine of both sides of the equation \(2\theta = 180^\circ - 3\theta\):

\(\sin(2\theta) = \sin(180^\circ - 3\theta)\)

Using the identity \(\sin(180^\circ - x) = \sin x\), we get:

\(\sin(2\theta) = \sin(3\theta)\)

Now, we use the double angle formula for sine (\(\sin(2\theta) = 2\sin\theta\cos\theta\)) and the triple angle formula for sine (\(\sin(3\theta) = 3\sin\theta - 4\sin^3\theta\)):

\(2\sin\theta\cos\theta = 3\sin\theta - 4\sin^3\theta\)

Since \(\theta = 36^\circ\), \(\sin\theta = \sin 36^\circ \ne 0\). We can divide both sides by \(\sin\theta\):

\(2\cos\theta = 3 - 4\sin^2\theta\)

Using the identity \(\sin^2\theta = 1 - \cos^2\theta\):

\(2\cos\theta = 3 - 4(1 - \cos^2\theta)\)

\(2\cos\theta = 3 - 4 + 4\cos^2\theta\)

Rearranging the terms to form a quadratic equation in terms of \(\cos\theta\):

\(4\cos^2\theta - 2\cos\theta - 1 = 0\)

Let \(x = \cos\theta\). The equation becomes \(4x^2 - 2x - 1 = 0\). We can solve this quadratic equation using the quadratic formula \(x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\):

\(x = \frac{-(-2) \pm \sqrt{(-2)^2 - 4(4)(-1)}}{2(4)}\)

\(x = \frac{2 \pm \sqrt{4 + 16}}{8}\)

\(x = \frac{2 \pm \sqrt{20}}{8}\)

\(x = \frac{2 \pm 2\sqrt{5}}{8}\)

\(x = \frac{1 \pm \sqrt{5}}{4}\)

So, \(\cos 36^\circ = \frac{1 \pm \sqrt{5}}{4}\). Since 36° is in the first quadrant, \(\cos 36^\circ\) must be positive. The value \(\frac{1 - \sqrt{5}}{4}\) is negative because \(1 < \sqrt{5}\). Therefore, we take the positive value:

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

Now we find \(\sin 36^\circ\) using the Pythagorean identity \(\sin^2\theta + \cos^2\theta = 1\):

\(\sin^2 36^\circ = 1 - \cos^2 36^\circ\)

\(\sin^2 36^\circ = 1 - \left(\frac{\sqrt{5}+1}{4}\right)^2\)

\(\sin^2 36^\circ = 1 - \frac{(\sqrt{5})^2 + 2(\sqrt{5})(1) + 1^2}{16}\)

\(\sin^2 36^\circ = 1 - \frac{5 + 2\sqrt{5} + 1}{16}\)

\(\sin^2 36^\circ = 1 - \frac{6 + 2\sqrt{5}}{16}\)

\(\sin^2 36^\circ = \frac{16 - (6 + 2\sqrt{5})}{16}\)

\(\sin^2 36^\circ = \frac{16 - 6 - 2\sqrt{5}}{16}\)

\(\sin^2 36^\circ = \frac{10 - 2\sqrt{5}}{16}\)

Taking the square root of both sides: \(\sin 36^\circ = \sqrt{\frac{10 - 2\sqrt{5}}{16}}\). Since 36° is in the first quadrant, \(\sin 36^\circ\) is positive, so we take the positive square root:

\(\sin 36^\circ = \frac{\sqrt{10 - 2\sqrt{5}}}{\sqrt{16}}\)

\(\sin 36^\circ = \frac{\sqrt{10 - 2\sqrt{5}}}{4}\)

Comparing this result with the given options, we find it matches the second option.

Revision Table: Key Trigonometric Values and Identities

Understanding special angle values and fundamental trigonometric identities is crucial for solving such problems. Here's a brief overview:

• Special Angles: 0°, 30°, 45°, 60°, 90°, 18°, 36°, 54°, 72°, etc., have specific derived trigonometric values.
• Pythagorean Identity: \(\sin^2\theta + \cos^2\theta = 1\)
• Double Angle Formulas: \(\sin(2\theta) = 2\sin\theta\cos\theta\), \(\cos(2\theta) = \cos^2\theta - \sin^2\theta = 2\cos^2\theta - 1 = 1 - 2\sin^2\theta\)
• Triple Angle Formulas: \(\sin(3\theta) = 3\sin\theta - 4\sin^3\theta\), \(\cos(3\theta) = 4\cos^3\theta - 3\cos\theta\)
• Angle Subtraction Identity: \(\sin(180^\circ - x) = \sin x\), \(\cos(180^\circ - x) = -\cos x\)

Additional Information: Relating Special Angles

The angles 18°, 36°, 54°, and 72° are often called special angles because their trigonometric values can be expressed using square roots and rational numbers. They are related to the properties of regular pentagons.

• Note that \(36^\circ = 90^\circ - 54^\circ\). So, \(\sin 36^\circ = \cos 54^\circ\) and \(\cos 36^\circ = \sin 54^\circ\).
• Also, \(18^\circ = 90^\circ - 72^\circ\). So, \(\sin 18^\circ = \cos 72^\circ\) and \(\cos 18^\circ = \sin 72^\circ\).
• Furthermore, \(36^\circ = 2 \times 18^\circ\). This allows deriving \(\sin 36^\circ\) and \(\cos 36^\circ\) using double angle formulas if \(\sin 18^\circ\) and \(\cos 18^\circ\) are known, or vice-versa. The value of \(\sin 18^\circ\) is \(\frac{\sqrt{5}-1}{4}\).