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

3.6 Trigonometric identities 127
Example3.5 Usethe( identities ) inTable 3.1( tosimplify ) the following expressions:
π 3π
(a) sin
2 + θ (b) cos
2 − θ
(c) tan(2π − θ) (d) sin(π − θ)
( π
)
Solution (a) The expression sin
2 + θ isof the form sin(A +B) and so we use the identity
sin(A +B) =sinAcosB +sinBcosA
PuttingA = π andB = θ weobtain
2
( π
)
sin
2 + θ = sin π 2 cos θ +sin θ cos π 2
We note thatsin π 2 = 1,cos π 2 =0andso
( π
)
sin
2 + θ =1cosθ +sinθ(0)
=cosθ
( ) 3π
(b) The expression cos
2 − θ has the form cos(A −B); hence we use the identity
cos(A −B) =cosAcosB +sinAsinB
PuttingA = 3π ,B = θ weobtain
2
( ) 3π
cos
2 − θ = cos 3π 2 cos θ +sin 3π 2 sinθ
Now cos 3π 2 = 0,sin 3π = −1 and so
2
( ) 3π
cos
2 − θ =0cosθ+(−1)sinθ
(c) We use the identity
= −sin θ
tan(A −B) = tanA−tanB
1+tanAtanB
SubstitutingA = 2π,B = θ we obtain
tan(2π − θ) =
tan2π −tan θ
1+tan2πtanθ
Since tan2π = 0 this simplifies to
tan(2π − θ) = −tanθ
1
(d) We use the identity
=−tanθ
sin(A −B) =sinAcosB −sinBcosA