082-Engineering-Mathematics-Anthony-Croft-Robert-Davison-Martin-Hargreaves-James-Flint-Edisi-5-2017

128 Chapter 3 The trigonometric functions
SubstitutingA = π,B = θ, this thenbecomes
sin(π−θ)=sinπcosθ −sinθcosπ
Now sinπ = 0,cos π = −1 and so we obtain
sin(π−θ)=0cosθ −sinθ(−1)=sinθ
Example3.6 Simplify
(a) cos(π + θ)
(b) tan(π − θ)
(c) sin 3 B +sinBcos 2 B
(d) tanA(1 +cos2A)
Solution (a) Usingthe identityfor cos(A +B) withA = π,B = θ weobtain
cos(π+θ)=cosπ cosθ −sinπ sinθ
= (−1)cosθ − (0)sinθ
= −cos θ
(b) Usingthe identityfor tan(A −B) withA = π,B = θ weobtain
tan(π−θ)= tanπ−tanθ
1+tanπtanθ
= 0−tanθ
1+(0)tanθ
=−tanθ
(c) sin 3 B +sinBcos 2 B =sinB(sin 2 B +cos 2 B)
=sinB since sin 2 B +cos 2 B = 1
(d) Firstlywe notethat tanA = sinA . Also wehave from Table 3.1
cosA
cos 2 A = 1+cos2A
2
from which
Hence
1+cos2A=2cos 2 A
tanA(1 +cos2A) = sinA
cosA 2cos2 A
=2sinAcosA
= sin2A