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Question

If sin2A+sin2B=12 and cos2A+cos2B=32, then the value of |cos(AB)| is

A
58
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B
38
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C
58
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D
38
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Solution

The correct option is A 58
sin2A+sin2B=12(1)
cos2A+cos2B=32(2)

Squaring and adding equations (1) and (2), we get
sin22A+sin22B+2sin2Asin2B+cos22A+cos22B+2cos2Acos2B=1+942+2(cos2Acos2B+sin2Asin2B)=522(1+cos2(AB))=52
4cos2(AB)=52
cos2(AB)=58|cos(AB)|=58

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