Inverse Trigonometric Functions - Finding Principle Value
Duration: 13 min
This video lesson is available to enrolled students.
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This educational video is a comprehensive lecture on finding the principal values of inverse trigonometric functions. The instructor begins by presenting a reference table that outlines the principal ranges and key identities for all six inverse trigonometric functions (sin⁻¹, cos⁻¹, tan⁻¹, cot⁻¹, sec⁻¹, csc⁻¹). The core of the lesson focuses on applying these identities to evaluate specific expressions. The instructor demonstrates the process for several examples, including sin⁻¹(-1/2), cos⁻¹(-1/2), cot⁻¹(-1/√3), and sec⁻¹(2/√3), using the formulas for negative arguments (e.g., sin⁻¹(-x) = -sin⁻¹(x)) and the identities for functions in the second quadrant (e.g., cos⁻¹(-x) = π - cos⁻¹(x)). The video concludes with a multiple-choice problem that combines these concepts, requiring the evaluation of tan⁻¹(√3) - sec⁻¹(-2), which is solved by applying the identities and known values. The entire lesson is delivered on a digital blackboard with clear, step-by-step calculations and explanations.
Chapters
0:00 – 2:00 00:00-02:00
The video opens with a title slide, "Finding Principal Values," and a comprehensive table of inverse trigonometric identities. The table lists the six functions (sin⁻¹, cos⁻¹, tan⁻¹, cot⁻¹, sec⁻¹, csc⁻¹) and their properties. For each function, it provides the principal range (e.g., [-π/2, π/2] for sin⁻¹) and the formula for negative arguments (e.g., sin⁻¹(-x) = -sin⁻¹(x) for the 4th quadrant). The instructor explains that these identities are used to find the principal value of an expression, which is the value within the principal range. The table also shows the identity for functions in the 2nd quadrant, such as cos⁻¹(-x) = π - cos⁻¹(x).
2:00 – 5:00 02:00-05:00
The instructor transitions to solving example problems. The first example is sin⁻¹(-1/2). The on-screen text shows the step-by-step solution: sin⁻¹(-1/2) = -sin⁻¹(1/2). The instructor explains that since sin⁻¹(1/2) = π/6, the result is -π/6. This is confirmed by noting that -π/6 is within the principal range of [-π/2, π/2]. The instructor then moves to the next example, cot⁻¹(-1/√3), and applies the identity cot⁻¹(-x) = π - cot⁻¹(x). The solution is shown as π - cot⁻¹(1/√3) = π - π/3 = 2π/3, which is within the principal range of (0, π).
5:00 – 10:00 05:00-10:00
The instructor continues with more examples. The next problem is cos⁻¹(-1/2). The solution is shown as cos⁻¹(-1/2) = π - cos⁻¹(1/2) = π - π/3 = 2π/3, which is within the principal range [0, π]. The video then shows the problem sec⁻¹(2/√3). The instructor explains that sec⁻¹(2/√3) = cos⁻¹(√3/2) = π/6, as the principal range of sec⁻¹ is [0, π] excluding π/2. The instructor then presents a multiple-choice question: tan⁻¹(√3) - sec⁻¹(-2). The solution is shown as π/3 - (π - sec⁻¹(2)) = π/3 - (π - π/3) = π/3 - 2π/3 = -π/3. The instructor notes that this is the correct answer, which is not listed, but the process is correct.
10:00 – 13:08 10:00-13:08
The instructor revisits the multiple-choice problem, tan⁻¹(√3) - sec⁻¹(-2). The on-screen text shows the calculation: tan⁻¹(√3) = π/3 and sec⁻¹(-2) = π - sec⁻¹(2) = π - π/3 = 2π/3. The final answer is π/3 - 2π/3 = -π/3. The instructor explains that this is the correct value, but it is not among the options (a) π, (b) -π/3, (c) π/3, (d) 2π/3. The instructor notes that option (b) is -π/3, which is the correct answer. The video concludes with a summary of the key identities and the process for finding principal values.
The video provides a structured and methodical approach to solving problems involving inverse trigonometric functions. It begins by establishing the fundamental identities and principal ranges, which are essential for determining the correct value. The core of the lesson is the application of these identities to evaluate expressions with negative arguments. The instructor demonstrates a clear, step-by-step process: identify the function, apply the appropriate identity (e.g., for negative input or for the second quadrant), substitute known values, and verify the result is within the principal range. The progression from simple examples to a more complex multiple-choice problem effectively reinforces the concepts. The key takeaway is that the principal value is not always the angle returned by a calculator, but the specific value within the defined range, and these identities are the tools to find it.