Let A and B (where A>B), be acute angles. If sin(A+B)=1213 and cos(A−B)=35, then the value(s) of sin(2A) is/are
A
5665
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B
1665
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C
−5665
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D
−1665
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Solution
The correct option is B1665 sin(A+B)=1213
So, A+B lies in first or second quadrant, ⇒cos(A+B)=±513 cos(A−B)=35 ∵A,B are acute angles and A>B⇒A−B lies in the first quadrant, ⇒sin(A−B)=45
Now, sin(2A)=sin(A+B+A−B)=sin(A+B)cos(A−B)+cos(A+B)sin(A−B)=1213×35±513×45=3665+2065 or 3665−2065=5665 or 1665