Question
Fri April 15, 2011 By: Chandan Tuli

trigonometry

Expert Reply
Thu May 05, 2011
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]
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