Calculus II ep11: Inverse trig functions (Oct 1, 2026)
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Overview
Inverse trigonometric functions return angles, but defining them as functions requires restricting sine and cosine to one-to-one intervals: arcsin uses sine on [-π/2, π/2], while arccos uses cosine on [0, π]. Prof Staecker uses right-triangle constructions to simplify compositions such as cos(arcsin x), then derives the arcsin and arccos derivatives by implicit differentiation and applies the chain and product rules to examples.
Key takeaways
- An inverse function exists only after restricting a non-one-to-one function: arcsin uses sine on [-π/2, π/2], and arccos uses cosine on [0, π].
- Arcsin and arccos both accept inputs in [-1, 1], but their output ranges differ: [-π/2, π/2] for arcsin and [0, π] for arccos.
- Right triangles derive compositions efficiently: cos(arcsin(x)) = √(1 - x²) and sin(arccos(x)) = √(1 - x²).
- Implicit differentiation gives d/dx arcsin(x) = 1/√(1 - x²) and d/dx arccos(x) = -1/√(1 - x²).
- When an inverse-trig derivative has a composite input, apply the chain rule separately; for arcsin(2x² + 7), the inner derivative contributes a factor of 4x.
Chapters
- Arcsin, arccos, and arctan are alternate names for inverse sine, cosine, and tangent.
- The notation sin⁻¹(x) means inverse sine, not 1/sin(x); the reciprocal is cosecant.
- For a right triangle with opposite side 1 and adjacent side 3, tan(θ) = 1/3, so θ = arctan(1/3).
- Reflecting the full sine graph across y = x produces a curve that fails the vertical line test.
- Sine is not one-to-one: for example, sin(π) and sin(2π) both equal 0.
- Restricting sine to a one-to-one interval makes its inverse a function.
- The principal arcsin branch uses sine on [-π/2, π/2], where sine is one-to-one.
- Arcsin has domain [-1, 1], since sine never takes values outside that interval.
- Its range is [-π/2, π/2]; for instance, arcsin(2) is undefined over the reals.
- Calculator output depends on whether the angle mode is radians or degrees: arcsin(1) is π/2 radians or 90°.
- Unit-circle values give arcsin(1/2) = π/6 and arccos(1/2) = π/3.
- Other standard values include arcsin(1) = π/2 and arcsin(-1/2) = -π/6.
- Set θ = arcsin(x), so sin(θ) = x; represent this as opposite side x and hypotenuse 1.
- The Pythagorean theorem gives the adjacent side √(1 - x²), so cos(arcsin(x)) = √(1 - x²).
- The same triangle yields tan(arcsin(x)) = x/√(1 - x²), without memorizing a separate identity.
- Arccos uses the one-to-one cosine interval [0, π], beginning at cos(0) = 1 and ending at cos(π) = -1.
- Its domain is [-1, 1], like arcsin, but its range is [0, π].
- Reflecting this restricted cosine graph across y = x gives the arccos graph, which differs in orientation and range from arcsin.
- Set θ = arccos(x), so cos(θ) = x; use adjacent side x and hypotenuse 1.
- The remaining side is √(1 - x²), giving sin(arccos(x)) = √(1 - x²).
- Taking opposite over adjacent gives tan(arccos(x)) = √(1 - x²)/x wherever the expression is defined.
- Starting with y = arcsin(x), rewrite the inverse relationship as x = sin(y).
- Implicit differentiation gives 1 = cos(y)·dy/dx, so dy/dx = 1/cos(y).
- Substituting y = arcsin(x) and using cos(arcsin(x)) = √(1 - x²) gives d/dx arcsin(x) = 1/√(1 - x²).
- The corresponding derivative is d/dx arccos(x) = -1/√(1 - x²); the negative sign comes from the derivative of cosine.
- For arcsin(2x² + 7), apply the arcsin derivative formula to the inside expression, then multiply by its derivative 4x.
- The chain-rule factor stays outside the inverse-trig derivative expression; it is not inserted under the radical.
- For arcsin(√x), the chain rule combines 1/√(1 - x) with the inner derivative 1/(2√x).
- For a product involving arccos(-x) and a polynomial factor, use the product rule and the arccos derivative.
- Differentiating arccos(-x) also requires the chain-rule factor -1 from the inner function; the next class will introduce arctan.
Summary, takeaways, and chapters were generated by AI from the video's transcript and may contain errors. The video belongs to its creator, Prof Staecker.