Question

Derive the value of sin 36

Answer 1

$\frac{\sqrt{10 - 2\sqrt{5}}}{4}$

Answer 2

To derive the value of $\sin 36^\circ$, we can follow these steps:

1. Set up the angle relationship: Let $\theta = 36^\circ$. Multiplying by 5, we get $5\theta = 180^\circ$. We can split $5\theta$ into $2\theta$ and $3\theta$: $2\theta + 3\theta = 180^\circ$ Rearranging the terms, we get: $2\theta = 180^\circ - 3\theta$

2. Apply the sine function to both sides: 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 $\sin(180^\circ - x) = \sin x$: $\sin(2\theta) = \sin(3\theta)$

3. Expand using double and triple angle formulas: Recall the formulas: $\sin(2\theta) = 2 \sin\theta \cos\theta$ $\sin(3\theta) = 3 \sin\theta - 4 \sin^3\theta$ Substitute these into the equation: $2 \sin\theta \cos\theta = 3 \sin\theta - 4 \sin^3\theta$

4. Solve for $\cos\theta$: Since $\theta = 36^\circ$, $\sin\theta = \sin 36^\circ \neq 0$. Therefore, we can divide both sides by $\sin\theta$: $2 \cos\theta = 3 - 4 \sin^2\theta$ Now, use the identity $\sin^2\theta = 1 - \cos^2\theta$: $2 \cos\theta = 3 - 4 (1 - \cos^2\theta)$ $2 \cos\theta = 3 - 4 + 4 \cos^2\theta$ $2 \cos\theta = -1 + 4 \cos^2\theta$ Rearrange the terms to form a quadratic equation in $\cos\theta$: $4 \cos^2\theta - 2 \cos\theta - 1 = 0$

Let $x = \cos\theta$. The equation becomes $4x^2 - 2x - 1 = 0$. Using the quadratic formula $x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$: $x = \frac{-(-2) \pm \sqrt{(-2)^2 - 4(4)(-1)}}{2(4)}$ $x = \frac{2 \pm \sqrt{4 + 16}}{8}$ $x = \frac{2 \pm \sqrt{20}}{8}$ $x = \frac{2 \pm 2\sqrt{5}}{8}$ $x = \frac{1 \pm \sqrt{5}}{4}$

Since $\theta = 36^\circ$ is in the first quadrant, $\cos 36^\circ$ must be positive. $\sqrt{5} \approx 2.236$. $\frac{1 + \sqrt{5}}{4} \approx \frac{1 + 2.236}{4} = \frac{3.236}{4} \approx 0.809$ (positive) $\frac{1 - \sqrt{5}}{4} \approx \frac{1 - 2.236}{4} = \frac{-1.236}{4} \approx -0.309$ (negative) Therefore, we choose the positive value: $\cos 36^\circ = \frac{1 + \sqrt{5}}{4}$

5. Calculate $\sin 36^\circ$: We know that $\sin^2\theta + \cos^2\theta = 1$. $\sin^2 36^\circ = 1 - \cos^2 36^\circ$ $\sin^2 36^\circ = 1 - \left(\frac{1 + \sqrt{5}}{4}\right)^2$ $\sin^2 36^\circ = 1 - \frac{(1)^2 + (\sqrt{5})^2 + 2(1)(\sqrt{5})}{16}$ $\sin^2 36^\circ = 1 - \frac{1 + 5 + 2\sqrt{5}}{16}$ $\sin^2 36^\circ = 1 - \frac{6 + 2\sqrt{5}}{16}$ To combine the terms, find a common denominator: $\sin^2 36^\circ = \frac{16}{16} - \frac{6 + 2\sqrt{5}}{16}$ $\sin^2 36^\circ = \frac{16 - (6 + 2\sqrt{5})}{16}$ $\sin^2 36^\circ = \frac{16 - 6 - 2\sqrt{5}}{16}$ $\sin^2 36^\circ = \frac{10 - 2\sqrt{5}}{16}$

Since $36^\circ$ is in the first quadrant, $\sin 36^\circ$ must be positive. $\sin 36^\circ = \sqrt{\frac{10 - 2\sqrt{5}}{16}}$ $\sin 36^\circ = \frac{\sqrt{10 - 2\sqrt{5}}}{4}$