Step 1: Understanding the Concept:
The problem asks for the ratio of two geometric areas. The first area is bounded by $y = \cos x$, the lines $x=0$, $x=\pi/3$, and the x-axis. The second area is bounded by $y = \cos 2x$ with the same boundary lines. We evaluate the area under each curve independently using definite integrals over the specified interval.
Step 2: Key Formula or Approach:
1. The area under a curve $y = f(x)$ above the x-axis from $a$ to $b$ is given by $A = \int_a^b f(x) dx$.
2. Standard integrations: $\int \cos(kx) dx = \frac{1}{k}\sin(kx)$.
Step 3: Detailed Explanation:
Let's calculate the area bounded by the first curve, $A_1$:
\[ A_1 = \int_0^{\frac{\pi}{3}} \cos x \,dx \]
Evaluate this definite integral:
\[ A_1 = [\sin x]_0^{\frac{\pi}{3}} = \sin\left(\frac{\pi}{3}\right) - \sin(0) \]
We know $\sin(60^\circ) = \frac{\sqrt{3}}{2}$:
\[ A_1 = \frac{\sqrt{3}}{2} - 0 = \frac{\sqrt{3}}{2} \]
Now, calculate the area bounded by the second curve, $A_2$:
\[ A_2 = \int_0^{\frac{\pi}{3}} \cos 2x \,dx \]
Evaluate this definite integral:
\[ A_2 = \left[\frac{\sin 2x}{2}\right]_0^{\frac{\pi}{3}} = \frac{\sin\left(2 \cdot \frac{\pi}{3}\right)}{2} - \frac{\sin(0)}{2} \]
\[ A_2 = \frac{\sin\left(\frac{2\pi}{3}\right)}{2} - 0 \]
Since $\sin(120^\circ) = \sin(180^\circ - 60^\circ) = \sin(60^\circ) = \frac{\sqrt{3}}{2}$:
\[ A_2 = \frac{\frac{\sqrt{3}}{2}}{2} = \frac{\sqrt{3}}{4} \]
Now, find the ratio $A_1 : A_2$:
\[ \text{Ratio} = \frac{A_1}{A_2} = \frac{\frac{\sqrt{3}}{2}}{\frac{\sqrt{3}}{4}} \]
\[ \text{Ratio} = \frac{\sqrt{3}}{2} \times \frac{4}{\sqrt{3}} = \frac{4}{2} = 2 \]
The ratio simplifies to $2 : 1$.
Step 4: Final Answer:
The ratio of the areas is $2 : 1$.