Gilbert Strang, Calculus, 3rd edition (Wellesley-Cambridge Press; MIT OpenCourseWare RES.18-001) §4.4 Inverses of Trigonometric Functions, #5 · Calculus · Differentiation rules

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We know that $sin π = 0$ . Why isn't $π = sin^{- 1} 0$ ?

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Gilbert Strang, Calculus, 3rd edition (Wellesley-Cambridge Press; MIT OpenCourseWare RES.18-001) · §4.4 Inverses of Trigonometric Functions, #5

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