“inverse trigonometric function”的概念、定义、翻译、参考文献-科学参考

inverse trigonometric function

Mathematics

Sine is not a one-to-one function, and so not invertible, but ‘inverse sine’ can be defined by choosing a set of principal values. Thus, if y = sin⁻¹ x, then x = sin y, but the converse need not hold. For each x in the interval [−1,1] there is a unique y in the interval [−π‎/2, π‎/2] such that x = sin y and this value of y is defined as y = sin⁻¹ x.

Similarly, for real x, y = tan⁻¹ x if x = tan y, and the unique value y is taken from the interval (−π‎/2, π‎/2). Also, for x in the interval [−1,1], y = cos⁻¹ x if x = cos y and the unique value y is taken from the interval [0, π‎].

The notation arcsin, arctan, and arccos for sin⁻¹, tan⁻¹, and cos⁻¹ is also used. Their derivatives are:

Graph of inverse sine

$\begin{array}{l} \frac{d}{d x} ({sin}^{- 1} x) = \frac{1}{\sqrt{1 - x^{2}}} (x \neq \pm 1), \\ \frac{d}{d x} ({cos}^{- 1} x) = - \frac{1}{\sqrt{1 - x^{2}}} (x \neq \pm 1), \\ \frac{d}{d x} ({tan}^{- 1} x) = \frac{1}{1 + x^{2}} . \end{array}$

Graph of inverse tangent

Graph of inverse cosine

单词	inverse trigonometric function
释义	inverse trigonometric function Mathematics Sine is not a one-to-one function, and so not invertible, but ‘inverse sine’ can be defined by choosing a set of principal values. Thus, if y = sin⁻¹ x, then x = sin y, but the converse need not hold. For each x in the interval [−1,1] there is a unique y in the interval [−π‎/2, π‎/2] such that x = sin y and this value of y is defined as y = sin⁻¹ x. Similarly, for real x, y = tan⁻¹ x if x = tan y, and the unique value y is taken from the interval (−π‎/2, π‎/2). Also, for x in the interval [−1,1], y = cos⁻¹ x if x = cos y and the unique value y is taken from the interval [0, π‎]. The notation arcsin, arctan, and arccos for sin⁻¹, tan⁻¹, and cos⁻¹ is also used. Their derivatives are: Graph of inverse sine $\begin{array}{l} \frac{d}{d x} ({sin}^{- 1} x) = \frac{1}{\sqrt{1 - x^{2}}} (x \neq \pm 1), \\ \frac{d}{d x} ({cos}^{- 1} x) = - \frac{1}{\sqrt{1 - x^{2}}} (x \neq \pm 1), \\ \frac{d}{d x} ({tan}^{- 1} x) = \frac{1}{1 + x^{2}} . \end{array}$ Graph of inverse tangent Graph of inverse cosine
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