Problem 3
Question
Find the exact value of each expression, if it is defined. \(\begin{array}{lll}{\text { (a) } \sin ^{-1} \frac{1}{2}} & {\text { (b) } \cos ^{-1}\left(-\frac{\sqrt{3}}{2}\right)} & {\text { (c) } \tan ^{-1}(-1)}\end{array}\)
Step-by-Step Solution
Verified Answer
(a) \(\frac{\pi}{6}\), (b) \(\frac{5\pi}{6}\), (c) \(-\frac{\pi}{4}\)
1Step 1: Understand the Problem
To find the exact value of the inverse trigonometric functions: (a) \(\sin^{-1}\left(\frac{1}{2}\right)\), (b) \(\cos^{-1}\left(-\frac{\sqrt{3}}{2}\right)\), and (c) \(\tan^{-1}(-1)\), we must identify angles whose trigonometric ratios correspond to the given values. These angles must be within the range of the respective inverse functions.
2Step 2: Solve for \(\sin^{-1}\left(\frac{1}{2}\right)\)
The inverse sine function, \(\sin^{-1}\), outputs angles in the range \([-\frac{\pi}{2}, \frac{\pi}{2}]\). We need an angle \(\theta\) such that \(\sin(\theta) = \frac{1}{2}\). This occurs at \(\theta = \frac{\pi}{6}\) or 30 degrees.
3Step 3: Solve for \(\cos^{-1}\left(-\frac{\sqrt{3}}{2}\right)\)
The inverse cosine function, \(\cos^{-1}\), outputs angles in the range \([0, \pi]\). We need an angle \(\theta\) such that \(\cos(\theta) = -\frac{\sqrt{3}}{2}\). This happens at \(\theta = \frac{5\pi}{6}\) or 150 degrees.
4Step 4: Solve for \(\tan^{-1}(-1)\)
The inverse tangent function, \(\tan^{-1}\), outputs angles in the range \((-\frac{\pi}{2}, \frac{\pi}{2})\). We need an angle \(\theta\) such that \(\tan(\theta) = -1\). This occurs at \(\theta = -\frac{\pi}{4}\) or -45 degrees.
Key Concepts
Understanding Trigonometric RatiosAngle Measurement in Inverse Trigonometric FunctionsFunction Ranges in Inverse Trigonometry
Understanding Trigonometric Ratios
Trigonometric ratios are fundamental in understanding angles and triangles.When dealing with right-angled triangles:
- Sine ( \( ext{sin}\)): This ratio is the length of the opposite side divided by the hypotenuse.
- Cosine ( \( ext{cos}\)): This ratio is the length of the adjacent side divided by the hypotenuse.
- Tangent ( \( ext{tan}\)): This ratio is the length of the opposite side divided by the adjacent side.
Angle Measurement in Inverse Trigonometric Functions
Inverse trigonometric functions return angle measurements, typically in radians or degrees. The importance of degrees and radians is highlighted when the inverse functions are used. It's beneficial to grasp the conversions:
- 360 degrees = 2π radians
- 180 degrees = π radians
- 90 degrees = π/2 radians
- \( ext{sin}^{-1}\) outputs angles in \([−\frac{\text{π}}{2},\frac{\text{π}}{2}]\) or [−90, 90] degrees.
- \( ext{cos}^{-1}\) outputs angles in \([0,\text{π}]\) or [0, 180] degrees.
- \( ext{tan}^{-1}\) outputs angles in \((-\frac{\text{π}}{2},\frac{\text{π}}{2})\) or [−90, 90) degrees.
Function Ranges in Inverse Trigonometry
The range of an inverse trigonometric function is the set of possible outputs, i.e., angles, it can produce. Each inverse function has its peculiar range due to the nature of the trigonometric ratio's relationship to the angle.
The inverse sine function, \( \sin^{-1} \), maps input ratios to angles within the range \([−\frac{\pi}{2},\frac{\pi}{2}]\). This means it finds angles only in the right half-circle, encompassing negative and positive outputs.
The \( \cos^{-1} \) function, conversely, returns angles within \([0, \pi]\). It covers the top half-circle, finding all relevant angles.Lastly, \( \tan^{-1} \) has the range \((-\frac{\pi}{2}, \frac{\pi}{2})\), covering half of the full circle range, with continuous values but excluding the boundaries.These ranges are set for convenience, ensuring that each input has one unique output, avoiding confusion over multiple possible angles that could have the same trigonometric ratio.
The inverse sine function, \( \sin^{-1} \), maps input ratios to angles within the range \([−\frac{\pi}{2},\frac{\pi}{2}]\). This means it finds angles only in the right half-circle, encompassing negative and positive outputs.
The \( \cos^{-1} \) function, conversely, returns angles within \([0, \pi]\). It covers the top half-circle, finding all relevant angles.Lastly, \( \tan^{-1} \) has the range \((-\frac{\pi}{2}, \frac{\pi}{2})\), covering half of the full circle range, with continuous values but excluding the boundaries.These ranges are set for convenience, ensuring that each input has one unique output, avoiding confusion over multiple possible angles that could have the same trigonometric ratio.
Other exercises in this chapter
Problem 2
The sign of a trigonometric function of \(\theta\) depends on the _____ in which the terminal side of the angle \(\theta\) lies. In Quadrant II, \(\sin \theta\)
View solution Problem 2
The reciprocal identities state that \(\csc \theta=\) \(\frac{1}{\square}\) \(\sec \theta=\) \(\frac{1}{\square}\) \(\cot \theta=\) \(\frac{1}{\square}\)
View solution Problem 3
Find the reference angle for the given angle. $$ \begin{array}{llll}{\text { (a) } 150^{\circ}} & {\text { (b) } 330^{\circ}} & {\text { (c) }-30^{\circ}}\end{a
View solution Problem 3
Find the radian measure of the angle with the given degree measure. $$ 72^{\circ} $$
View solution