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 Rough 3𝜋/5 = (3 × 180)/5 = 108° 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