Example 9 - Chapter 2 Class 12 Inverse Trigonometric Functions

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Example, 9 Find the value of sin-1 (sin 3π/5) Let y = sin-1(sin 3π/5) sin y = sin (3π/5) sin y = sin (108°) But, range of principal value of sin-1 is [(−π)/2, π/2] i.e. [-90° ,90° ] Hence y = 108° not possible Now, sin y = sin (108°) sin y = sin (180° – 72°) sin y = sin (72°) sin y = sin (72 × 𝜋/180) sin y = sin 2𝜋/5 Hence, y = 2𝜋/5 Which is in the range of principal value of sin-1 i.e. [(−π)/2, π/2] Hence, sin-1(sin 3π/5) = y = 2𝜋/5