Question

The value of $\sin 18^{\circ}$ is

• A. 5 - 15
• B. 5 - 14
• C. 5 + 14
• D. 5 + 15

Hint:

$\sin 18^{\circ}$ is a very famous value in coordinate geometry and trigonometry. Memorizing $\sin 18^{\circ} = \frac{\sqrt{5}-1}{4}$ along with its partner $\cos 36^{\circ} = \frac{\sqrt{5}+1}{4}$ helps bypass long quadratic derivations during tests!

Answer 1

The Correct Option is B

Step 1: Set up a useful equation.
Let $A=18^\circ$. Then $5A=90^\circ$, which we can split as $2A=90^\circ-3A$.

Step 2: Take sine of both sides.
$\sin2A=\sin(90^\circ-3A)=\cos3A$. Expanding, $2\sin A\cos A=4\cos^3A-3\cos A$.

Step 3: Cancel $\cos A$ and simplify.
Since $\cos18^\circ\neq0$, divide it out: $2\sin A=4\cos^2A-3$. Replace $\cos^2A=1-\sin^2A$ to get $4\sin^2A+2\sin A-1=0$.

Step 4: Solve the quadratic.
\[ \sin A=\frac{-2\pm\sqrt{4+16}}{8}=\frac{-1\pm\sqrt5}{4} \] Since $18^\circ$ is in the first quadrant, the sine is positive. \[ \boxed{\sin18^\circ=\frac{\sqrt5-1}{4}} \]