prove that cos^2 A+ cos^2 B- 2cosAcosBcos(A+B)= sin^2 (A+B)
Asked by
| 30th May, 2012,
04:42: PM
Expert Answer:
LHS: cos2 A + cos2B -2cosAcosBcos(A+B)
= cos2 A+ cos2B-2cosAcosB(cosAcosB-sinAsinB)
=cos2 A+ cos2B-2cos2Acos2B +2cosAcosBsinAsinB
=cos2 A -cos2Acos2B+ cos2B- cos2Acos2B +2cosAcosBsinAsinB
=cos2 A (1-cos2B) +cos2B (1-cos2 A) +2cosAcosBsinAsinB
=cos2 A sin2 B + cos2B sin2A + 2cosAcosBsinAsinB
= sin2 (A+B) = RHS
LHS: cos2 A + cos2B -2cosAcosBcos(A+B)
= cos2 A+ cos2B-2cosAcosB(cosAcosB-sinAsinB)
=cos2 A+ cos2B-2cos2Acos2B +2cosAcosBsinAsinB
=cos2 A -cos2Acos2B+ cos2B- cos2Acos2B +2cosAcosBsinAsinB
=cos2 A (1-cos2B) +cos2B (1-cos2 A) +2cosAcosBsinAsinB
=cos2 A sin2 B + cos2B sin2A + 2cosAcosBsinAsinB
= sin2 (A+B) = RHS
Answered by
| 31st May, 2012,
10:36: AM
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