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

130 Chapter 3 The trigonometric functions
Hence
sin70 ◦ −sin30 ◦
= 2sin20◦ cos50 ◦
cos50 ◦ cos50 ◦
= 2sin20 ◦
EXERCISES3.6
1 Usethe identities forsin(A ±B),cos(A ±B)and
tan(A ±B)to ( simplifythe ) following: ( )
(a) sin
θ − π 2
(b) cos
θ − π 2
(c) tan(θ + π) (d) sin(θ − π)
(e) cos(θ − π)
(g) sin(θ + π)
(
)
(i) sin 2θ + 3π 2
( )
π
(k) cos
2 + θ
(f) tan(θ −3π)
( )
(h) cos θ + 3π 2
( )
(j) cos
θ − 3π 2
2 Writedown the trigonometricidentityfortan(A
( )
+ θ).
BylettingA → π 2 showthat tan π
2 + θ can be
simplified to −cot θ.
3 (a) Bydividing the identity
sin 2 A +cos 2 A = 1 bycos 2 A showthat
tan 2 A +1 = sec 2 A.
(b) Bydividing the identity
sin 2 A +cos 2 A = 1 bysin 2 A showthat
1 +cot 2 A = cosec 2 A.
4 Simplify the followingexpressions:
(a) cosAtanA (b) sin θ cot θ
(c) tanB cosecB
( )
(d) cot2xsec2x
π
(e) tanθ tan
2 + θ (f) sin2t
cost
[Hint: seeQuestion 2.]
(g) sin 2 A +2cos 2 A
(h) 2cos 2 B −1
(i) (1 +cot 2 X)tan 2 X (j) (sin 2 A +cos 2 A) 2
(k) 1 2 sin2AtanA
(m)
sin2A
cos2A
(o) (tan 2 θ +1)cot 2 θ
5 Simplify
(a) sin110 ◦ −sin70 ◦
(b) cos20 ◦ −cos80 ◦
(c) sin40 ◦ +sin20 ◦
(d)
6 Show that
cos50 ◦ +cos40 ◦
√
2
sin60 ◦ +sin30 ◦
sin50 ◦ −sin40 ◦
isequivalent to
cos15 ◦
sin5 ◦
(l) (sec2 t −1)cos 2 t
(n)
sinA
sin2A
(p) cos2A +2sin 2 A
Solutions
1 (a) −cos θ (b) sin θ (c) tan θ
(d) −sin θ (e) −cos θ (f) tan θ
(g) −sin θ (h) sin θ (i) −cos2θ
(j) −sin θ (k) −sin θ
4 (a) sinA (b) cos θ (c) secB
(d) cosec 2x (e) −1 (f) 2sint
(g) 1 +cos 2 A (h) cos2B (i) sec 2 X
(j) 1 (k) sin 2 A (l) sin 2 t
(m) tan2A
(p) 1
(n)
1
2 secA (o) cosec 2 θ
5 (a) 0 (b) sin50 ◦ (c) cos10 ◦ (d) cos5 ◦