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Figure 6.13Proof of sin(A+B)=sinAcosB+cosAsinB.
PuttingA=B in the compound angle identities immediately gives thedouble angle
formulae:
sin 2A≡sinAcosA
cos 2A≡cos^2 −sin^2 A
≡2cos^2 A− 1
≡ 1 −2sin^2 A
tanA≡
2tanA
1 −tan^2 A
The cos 2Aresults are often more useful in the form
sin^2 A=^12 ( 1 −cos 2A)
cos^2 A=^12 ( 1 +cos 2A)
From the double angle formulae we easily deduce thehalf angle identities.Ift=tanθ/2,
then:
tanθ≡
2 t
1 −t^2
sinθ≡
2 t
1 +t^2
cosθ≡
1 −t^2
1 +t^2
For example:
tanθ=tan( 2 ×θ/ 2 )=
2tanθ/ 2
1 −tan^2 θ/ 2
=
2 t
1 −t^2
from the double angle formula.
Solution to review question 6.1.7
A.sin(A+B)=sinAcosB+sinBcosAis so important it should be at
your fingertips.