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`Sin^-1(Sin (17pi)/8)` - Mathematics

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प्रश्न

`sin^-1(sin  (17pi)/8)`

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उत्तर

We know

`sin(sin^-1theta)=theta if - pi/2<=theta<=pi/2`

We have

`sin^-1(sin  (17pi)/8)=sin^-1{sin(2pi+pi/8)}`

`=sin^-1(sin  pi/8)`

`=pi/8`

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पाठ 4: Inverse Trigonometric Functions - Exercise 4.07 [पृष्ठ ४२]

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आरडी शर्मा Mathematics [English] Class 12
पाठ 4 Inverse Trigonometric Functions
Exercise 4.07 | Q 1.05 | पृष्ठ ४२

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