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Trignometry

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10 Jan 2008 23:58:51 IST

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find the angle

Engineering Entrance , JEE Main , JEE Advanced , Mathematics , Trigonometry

find the angle of sin^-1(sin 5) please help me

Comments (5)

Aditya Arora

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11 Jan 2008 00:03:29 IST

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The answer is 5 radians

Rohit Gupta

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11 Jan 2008 18:48:02 IST

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Sneha,

sin^-1x can be max pi/2 and minimum -pi/2 (as far as i remember the range of sin^-1x)

let A = sin^-1(sin 5)
=> sin A = sin 5 = sin (pi + (5 - pi)) = - sin (5 - pi)
=> sin A = - sin (pi/2 + (5 - 3*pi/2)) = - cos (5 - 3*pi/2)

=> - sin A = cos (5 - 3*pi/2)

=> cos (pi/2 + A) = cos (5 - 3*pi/2)

=> pi/2 + A = 5 - 3*pi/2

=> A = (5 - 2*pi) is the correct answer.

(remember that answer must lie between -pi/2 and pi/2, and (5 - 2*pi) does)
.

Eshwar bandaru

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18 Jan 2008 17:17:11 IST

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3pi/4<5<2pi
Range of sininverse(sinx) is ? pi/ to +pi/2
Sin(2pi-5)= -sin5 ---------1
Sininverse(sin(2pi-5)) = 2pi-5 as 2pi-5is in the range
BUT 2pi-5= -sininverse(sin5)
Therefore ans=5-2pi

jacqueline massacky

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19 Jan 2008 03:50:56 IST

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the answer is 5

ashwin kumar

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19 Jan 2008 18:55:30 IST

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this is rsa,
range of sin-1 x = -pi/2 to pi/2

sin A = sin 5 = sin (pi + (5 - pi)) = - sin (5 - pi)
sin A = - sin (pi/2 + (5 - 3*pi/2)) = - cos (5 - 3*pi/2)
- sin A = cos (5 - 3*pi/2)
cos (pi/2 + A) = cos (5 - 3*pi/2)
pi/2 + A = 5 - 3*pi/2

A = 5-2pi

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