CBSE Class 12-science Answered
trigonometry
Asked by chandant | 15 Apr, 2011, 06:46: AM
Expert Answer
The question which you have posted is wrong. The correct statement of the question is as follows:
3sin2A+ 2sin2B =1 ... (1)
and 3 sin2A -2sin2B =0 ... (2)
Following is the solution:
From equation (1), we have:
3sin2A = 1 - 2sin2B = cos 2B ... (3)
From (2), we have:
3 sin2A = 2 sin 2B ... (4)
Consider cos (A + 2B) = cosA cos 2B - sin A sin 2B
= cosA (3sin2A) - sin A (3 sin 2A/2) [Using (3) and (4)]
= 3 cosA sin2A - sin A (3 sin A cosA) [Since, sin 2A = 2 sinA cos A]
= 3 cosA sin2A - 3 sin2 A cosA
= 0
Thus, A + 2B = ?/ 2
[Since, cos ?/ 2 = 0]
Answered by | 05 May, 2011, 10:44: AM
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