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Evaluate cos[cos-1(-32)+π6]

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प्रश्न

Evaluate `cos[cos^-1 ((-sqrt(3))/2) + pi/6]`

योग
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उत्तर

`cos[cos^-1 ((-sqrt(3))/2) + pi/6] = cos[cos^-1(cos  (5pi)/6) + pi/6]`   ......`(because  cos  (5pi)/6 = (-sqrt(3))/2)`

= `cos((5pi)/6 + pi/6)`  ......`(because cos^-1 cosx = x; x ∈ [0, pi])`

= `cos(pi)` = –1

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अध्याय 2: Inverse Trigonometric Functions - Exercise [पृष्ठ ३५]

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एनसीईआरटी एक्झांप्लर Mathematics Exemplar [English] Class 12
अध्याय 2 Inverse Trigonometric Functions
Exercise | Q 2 | पृष्ठ ३५

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