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Example 9 -  Find value of cos (-1710) - Chapter 3 Class 11

Example 9 - Chapter 3 Class 11 Trigonometric Functions - Part 2

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Example 9 Find the value of cos (–1710°). cos⁡(−1710°) = cos⁡(1710°) = cos⁡(1710 ×𝜋/180) = cos⁡(19𝜋/2) = cos⁡(9 1/2 𝜋 ) = cos⁡(10𝜋−𝜋/2) Value of cos x repeats after an interval of 2π, Hence, ignoring 5 × (2π) i.e. 10π = cos⁡((−𝜋)/2) = cos⁡((−180)/2) = cos⁡(–90°) = cos⁡90° = 0 As cos (−x) = cos x

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