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What is the value of the expression cos2Acos2B + sin2(A - B) - sin2(A + B)?
Answer & Solution
Correct Answer:
Option
C
cos2Acos2B + sin2(A - B) - sin2(A + B)
= cos2Acos2B + [{sin(A - B) + sin(A + B)}{sin(A - B) - sin(A + B)}]
= cos2Acos2B + [(sinAcosB - cosAsinB + sinAcosB + cosAsinB)(sinAcosB - cosAsinB - sinAcosB - cosAsinB)]
= cos2Acos2B + [(2sinAcosB) × (-2cosAsinB)]
= cos2Acos2B - (2sinAcosA) × (2sinBcosB)
= cos2Acos2B - sin2Asin2B
= cos(2A + 2B)
Alternate:
cos2Acos2B + sin2(A - B) - sin2(A + B)
= cos2Acos2B + sin2Asin2B
= cos(2A + 2B)
= cos2Acos2B + [{sin(A - B) + sin(A + B)}{sin(A - B) - sin(A + B)}]
= cos2Acos2B + [(sinAcosB - cosAsinB + sinAcosB + cosAsinB)(sinAcosB - cosAsinB - sinAcosB - cosAsinB)]
= cos2Acos2B + [(2sinAcosB) × (-2cosAsinB)]
= cos2Acos2B - (2sinAcosA) × (2sinBcosB)
= cos2Acos2B - sin2Asin2B
= cos(2A + 2B)
Alternate:
cos2Acos2B + sin2(A - B) - sin2(A + B)
= cos2Acos2B + sin2Asin2B
= cos(2A + 2B)
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