CBSE Class 12-science Answered
the value of sin[(1/4)*(arcsin(?63/8))]is?
Asked by Rishabh Jaiswal | 26 Jul, 2013, 09:23: AM
Expert Answer
Let sin^-1 (?63)/8 = ?
=> sin? = (?63)/8
=> cos? = ?(1 - 63/64) = 1/8
sin{(1/4)*sin^-1[(?(63))/8]}
= sin(?/4)
= ?[(1 - cos(?/2))/2] ... ( 1 )
cos(?/2)
= ?[(1 + cos?)/2]
= ?[(1 + 1/8)/2]
= 3/4
Putting this value of cos(?/2) = 3/4 in ( 1 ),
sin{(1/4)*sin^-1[(?(63))/8]}
= ?[(1 - 3/4)/2]
= 1/(2?2)
=> sin? = (?63)/8
=> cos? = ?(1 - 63/64) = 1/8
sin{(1/4)*sin^-1[(?(63))/8]}
= sin(?/4)
= ?[(1 - cos(?/2))/2] ... ( 1 )
cos(?/2)
= ?[(1 + cos?)/2]
= ?[(1 + 1/8)/2]
= 3/4
Putting this value of cos(?/2) = 3/4 in ( 1 ),
sin{(1/4)*sin^-1[(?(63))/8]}
= ?[(1 - 3/4)/2]
= 1/(2?2)
Answered by | 28 Jul, 2013, 02:06: PM
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