Finding the Principal Value of sin inverse sin(2π/3)

The question asks for the principal value of the expression \(\sin^{-1} \left( \sin \dfrac{2 \pi}{3} \right)\). To solve this, we need to understand the concept of the principal value branch of the inverse sine function, also known as sin inverse. The function \(\sin^{-1}(x)\) gives the angle \(\theta\) such that \(\sin(\theta) = x\).

Understanding the Principal Value Branch of sin inverse

The sine function, \(\sin(x)\), is periodic and not one-to-one over its entire domain. To define its inverse function, we restrict the domain of \(\sin(x)\) to an interval where it is one-to-one and covers its entire range [-1, 1]. This restricted interval is \([-\pi/2, \pi/2]\).

• The domain of \(\sin^{-1}(x)\) is \([-1, 1]\).
• The principal value branch of the range of \(\sin^{-1}(x)\) is \([-\pi/2, \pi/2]\).

Therefore, for \(\sin^{-1}(\sin \theta)\) to be equal to \(\theta\), the angle \(\theta\) must lie within the principal value range \([-\pi/2, \pi/2]\).

Evaluating sin(2π/3)

The given angle is \(2\pi/3\). Let's first find the value of \(\sin(2\pi/3)\). The angle \(2\pi/3\) is in the second quadrant.

We can write \(2\pi/3\) as \(\pi - \pi/3\).

Using the property \(\sin(\pi - \theta) = \sin \theta\), we have:

\[ \sin \left( \dfrac{2 \pi}{3} \right) = \sin \left( \pi - \dfrac{\pi}{3} \right) = \sin \left( \dfrac{\pi}{3} \right) \]

We know that \(\sin(\pi/3) = \sqrt{3}/2\).

So, the original expression becomes \(\sin^{-1} \left( \sin \dfrac{\pi}{3} \right)\).

Finding the Principal Value

Now we need to find the principal value of \(\sin^{-1} \left( \sin \dfrac{\pi}{3} \right)\).

We have the expression \(\sin^{-1}(\sin \theta)\) where \(\theta = \pi/3\). We check if this value of \(\theta\) is within the principal value range of \(\sin^{-1}(x)\), which is \([-\pi/2, \pi/2]\).

• \( \pi/3 \) is approximately \( 3.14 / 3 \approx 1.047 \) radians.
• \( -\pi/2 \) is approximately \( -3.14 / 2 = -1.57 \) radians.
• \( \pi/2 \) is approximately \( 3.14 / 2 = 1.57 \) radians.

Clearly, \( -\pi/2 \le \pi/3 \le \pi/2 \). The angle \(\pi/3\) lies within the principal value branch \([-\pi/2, \pi/2]\).

Since \(\pi/3\) is in the principal value range, we can use the identity \(\sin^{-1}(\sin \theta) = \theta\).

Therefore, the principal value of \(\sin^{-1} \left( \sin \dfrac{2 \pi}{3} \right)\) is:

\[ \sin^{-1} \left( \sin \dfrac{2 \pi}{3} \right) = \sin^{-1} \left( \sin \dfrac{\pi}{3} \right) = \dfrac{\pi}{3} \]

Summary of Steps

1. Identify the angle inside the sine function: \(2\pi/3\).
2. Check if this angle is within the principal value range of \(\sin^{-1}\) (which is \([-\pi/2, \pi/2]\)). Since \(2\pi/3\) is not in this range, we cannot simply say the answer is \(2\pi/3\).
3. Evaluate \(\sin(2\pi/3)\). We found \(\sin(2\pi/3) = \sin(\pi/3)\).
4. The expression becomes \(\sin^{-1}(\sin(\pi/3))\).
5. Check if the new angle \(\pi/3\) is within the principal value range \([-\pi/2, \pi/2]\). Yes, it is.
6. Since \(\pi/3\) is in the range, the principal value of \(\sin^{-1}(\sin(\pi/3))\) is \(\pi/3\).

Thus, the principal value of sin inverse sin(2π/3) is \(\pi/3\).

This detailed explanation shows how to find the principal value correctly by considering the range of the sin inverse function.