The Student Room Group

C3 Trig identities

In the text book there is the identity:
2sinAsinB[cos(A+B)cos(AB)]2sin A sin B \equiv -[cos (A+B)-cos (A-B)]

However on the sheet my teacher has given me it is written as:

2sinAsinBcos(AB)cos(AB)2sin A sin B \equiv cos (A-B)-cos (A-B)

Is the second still correct? Wouldn't cos(AB)cos(AB)=0cos (A-B)-cos (A-B)=0 ??
(edited 11 years ago)
Reply 1
Original post by Random999
Wouldn't cos(AB)cos(AB)=0cos (A-B)-cos (A-B)=0 ??


Yes



2sinAsinBcos(AB)cos(AB)2sin A sin B \equiv cos (A-B)-cos (A-B)




Are you sure that your teacher did not give you

2sinAsinBcos(AB)cos(A+B)2sin A sin B \equiv cos (A-B)-cos (A+B)
Reply 2
Original post by TenOfThem

Are you sure that your teacher did not give you

2sinAsinBcos(AB)cos(A+B)2sin A sin B \equiv cos (A-B)-cos (A+B)


This is not what I was given, however it was probably a typo on the sheet.

Thanks for your help. :smile:
Reply 3
Mistake in copying out?

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