InterviewSolution
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InterviewSolution
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ज्ञात करें `sin[(pi)/(2) - sin^(-1)(-(sqrt(3))/(2))]` |
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Answer» माना कि `sin^(-1)(-(sqrt(3))/(2)) = theta,` तो `theta in[(-pi)/(2),(pi)/(2)]` साथ ही `sin theta = (-sqrt(3))/(2) = - sin""(pi)/(3) = sin(-(pi)/(3)) rArr = - (pi)/(3) rArr sin ^(-1)((-sqrt(3))/(2)) = - (pi)/(3)` अब `sin[(pi)/(2) - sin^(-1)((-sqrt(3))/(2))] = sin[(pi)/(2) - ((-pi)/(3))] = sin ""(5pi)/(6) = (1)/(2)` |
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