Problem 38

Question

\(23-44=\) Find the exact value of the expression, if it is defined. \(\sin ^{-1}\left(\sin \left(\frac{11 \pi}{4}\right)\right)\)

Step-by-Step Solution

Verified
Answer
\( \sin ^{-1} \left( \sin \left(\frac{11\pi}{4} \right) \right) = \frac{\pi}{4} \).
1Step 1: Convert angle to standard range
First, we need to convert the angle \( \frac{11 \pi}{4} \) to an angle within the standard range \( [0, 2\pi] \). We do this by subtracting \( 2\pi \), which is equivalent to \( \frac{8\pi}{4} \), from \( \frac{11\pi}{4} \) to get \( \frac{3\pi}{4} \). Thus, \( \sin \left(\frac{11\pi}{4}\right) = \sin \left(\frac{3\pi}{4}\right) \).
2Step 2: Find the sine value of converted angle
Next, we calculate \( \sin \left(\frac{3\pi}{4}\right) \). Since \( \frac{3\pi}{4} \) is in the second quadrant, where sine is positive and is equal to \( \sin(\pi - \frac{\pi}{4}) \), the sine value is \( \sin \left(\frac{\pi}{4}\right) = \frac{\sqrt{2}}{2} \).
3Step 3: Apply inverse sine to the sine value
Finally, apply the inverse sine function, \( \sin^{-1} \), to \( \frac{\sqrt{2}}{2} \). The angle whose sine is \( \frac{\sqrt{2}}{2} \) that lies within the principal range of \( \sin^{-1} \), \([-\frac{\pi}{2}, \frac{\pi}{2}]\), is \( \frac{\pi}{4} \).

Key Concepts

Angle ConversionStandard Angle RangeInverse Sine FunctionTrigonometric Identities
Angle Conversion
Understanding angle conversion is crucial in trigonometry, especially when dealing with the unit circle and periodic functions. Angles expressed in radians can often be greater than the full circle range, which is from \(0\) to \(2\pi\), necessitating conversion into this range. In our problem, we have the angle \(\frac{11\pi}{4}\), which is beyond \(2\pi\).
To convert this angle into the standard range \([0, 2\pi]\), we subtract \(2\pi\) from \(\frac{11\pi}{4}\). This is equivalent to subtracting \(\frac{8\pi}{4}\) from \(\frac{11\pi}{4}\), resulting in \(\frac{3\pi}{4}\). Effectively, this means finding an angle on the unit circle that coincides with the given angle, hence making calculations manageable.
Standard Angle Range
The standard angle range in trigonometry is an essential concept for simplifying problems. It restricts the angles to within one full revolution of the circle, making calculations more straightforward. The typical standard range is from \(0\) to \(2\pi\) radians.
By converting angles like \(\frac{11\pi}{4}\) into this range, we ensure consistency in trigonometric functions. Our result, \(\frac{3\pi}{4}\), fits within this interval and allows us to use known trigonometric identities and values. This conversion is vital for simplifying expressions and ensuring that our results are meaningful and interpretable.
Inverse Sine Function
The inverse sine function, denoted as \(\sin^{-1}\), is used to find angles whose sine is a given number. Its range is particularly significant, returning angles only between \(-\frac{\pi}{2}\) and \(\frac{\pi}{2}\). This ensures that each value within this range corresponds uniquely to a single angle.
When we compute \(\sin^{-1}\left(\sin\left(\frac{3\pi}{4}\right)\right)\), we determine the angle in the function's principal range whose sine is \(\frac{\sqrt{2}}{2}\). This specific angle is \(\frac{\pi}{4}\), since \(\frac{3\pi}{4}\) can be expressed as \(\pi - \frac{\pi}{4}\) meaning that its sine value is indeed \(\frac{\sqrt{2}}{2}\).
The careful choice of range ensures the inverse function correctly identifies the appropriate angle.
Trigonometric Identities
Trigonometric identities simplify and solve expressions involving angles; they are core tools in trigonometric calculations. Here, we use the identity for sine in different quadrants.
For our converted angle \(\frac{3\pi}{4}\), which lies in the second quadrant, we use the identity \(\sin(\pi - x) = \sin(x)\). This means in the second quadrant, sine values mirror those in the first quadrant but remain positive, simplifying \(\sin\left(\frac{3\pi}{4}\right)\) to \(\frac{\sqrt{2}}{2}\) due to \(\sin\left(\frac{\pi}{4}\right)\).
Understanding how these identities work helps in effectively converting and dealing with trigonometric expressions. They are the backbone of manipulating trigonometric equations, leading to simplified solutions.