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Example 9 - Chapter 2 Inverse trigonometry - Find sin-1 (sin

Example 9 - Chapter 2 Class 12 Inverse Trigonometric Functions - Part 2

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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 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 𝟐𝝅/𝟓 Hence, y = 2𝜋/5 Which is in the range of sin-1 i.e. [(−π)/2, π/2] Hence, sin-1("sin " 3π/5) = y = 𝟐𝝅/𝟓

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Davneet Singh has done his B.Tech from Indian Institute of Technology, Kanpur. He has been teaching from the past 12 years. He provides courses for Maths, Science, Social Science, Physics, Chemistry, Computer Science at Teachoo.