170 Higher Engineering Mathematics
LHS =
1 − cos 2θ
sin 2θ
=
1 − (1 − 2 sin 2 θ)
2 sin θ cos θ
=
2 sin 2 θ
2 sin θ cos θ
=
sin θ
cos θ
= tan θ = RHS
Problem 13. Prove that
cot 2x + cosec 2x = cot x.
LHS = cot 2x + cosec 2x =
cos 2x
sin 2x
+
1
sin 2x
=
cos 2x + 1
sin 2x
=
(2 cos 2 x − 1) + 1
sin 2x
=
2 cos 2 x
sin 2x
=
2 cos 2 x
2 sin x cos x
=
cos x
sin x
= cot x = RHS
Problem 14. Solve the equation
cos 2θ + 3 sinθ = 2 for θ in the range 0 ◦ ≤ θ ≤ 360 ◦ .
Replacing the double angle term with the relationship
cos 2θ = 1 − 2 sin
2
θ gives:
1 − 2 sin
2
θ + 3 sinθ = 2
Rearranging gives:
−2 sin 2 θ + 3 sin θ − 1 = 0
or
2 sin 2 θ − 3 sin θ + 1 = 0
which is a quadratic in sin θ
Using the quadratic formula or by factorising gives:
(2 sinθ − 1)(sin θ − 1) = 0
from which, 2 sinθ − 1 = 0 or sinθ − 1 = 0
and
sin θ =
1
2 or sin θ = 1
from which,
θ = 30
◦ or 150
◦ or 90
◦
Now try the following exercise
Exercise 74 Further problems on double
angles
1. The power p in an electrical circuit is given by
p =
v 2
R
. Determine the power in terms of V ,
R and cos 2t when v = V cos t .
V
2
2R
(1 + cos 2t )
2. Prove the following identities:
(a) 1 −
cos 2φ
cos 2 φ
= tan 2 φ
(b)
1 + cos 2t
sin 2 t
= 2 cot 2 t
(c)
(tan 2x)(1 + tan x)
tan x
=
2
1 − tan x
(d) 2 cosec 2θ cos 2θ = cot θ − tan θ
3. If the third harmonic of a waveform is given by
V 3 cos 3θ, express the third harmonic in terms
of the first harmonic cosθ, when V 3 = 1.
[cos 3θ = 4 cos 3 θ − 3 cosθ]
In Problems 4 to 8, solve for θ in the range
−180 ◦ ≤ θ ≤ 180 ◦
4. cos 2θ = sin θ
[−90 ◦ , 30 ◦ , 150 ◦ ]
5. 3 sin 2θ + 2 cosθ = 0
[−160.47 ◦ , −90 ◦ , −19.47 ◦ , 90 ◦ ]
6. sin 2θ + cos θ = 0
[−150 ◦ , −90 ◦ , −30 ◦ , 90 ◦ ]
7. cos 2θ + 2 sin θ = −3
[ −90 ◦ ]
8. tan θ + cot θ = 2
[ 4 5 ◦ , −135 ◦ ]
17.4 Changing products of sines and
cosines into sums or differences
(i) sin(A + B) + sin(A − B) = 2 sin A cos B (from the
formulae in Section 17.1)
i.e. sin A cos B
=
1
2 [sin(A + B) + sin(A − B)]
(1)
(ii) sin(A + B) − sin(A − B) = 2 cos A sin B
i.e. cos A sin B
=
1
2 [sin(A + B) − sin(A − B)]
(2)
(iii) cos(A + B) + cos(A − B) = 2 cos A cos B
i.e. cos A cos B
=
1
2 [cos(A + B) + cos(A − B)]
(3)
LHS =
1 − cos 2θ
sin 2θ
=
1 − (1 − 2 sin 2 θ)
2 sin θ cos θ
=
2 sin 2 θ
2 sin θ cos θ
=
sin θ
cos θ
= tan θ = RHS
Problem 13. Prove that
cot 2x + cosec 2x = cot x.
LHS = cot 2x + cosec 2x =
cos 2x
sin 2x
+
1
sin 2x
=
cos 2x + 1
sin 2x
=
(2 cos 2 x − 1) + 1
sin 2x
=
2 cos 2 x
sin 2x
=
2 cos 2 x
2 sin x cos x
=
cos x
sin x
= cot x = RHS
Problem 14. Solve the equation
cos 2θ + 3 sinθ = 2 for θ in the range 0 ◦ ≤ θ ≤ 360 ◦ .
Replacing the double angle term with the relationship
cos 2θ = 1 − 2 sin
2
θ gives:
1 − 2 sin
2
θ + 3 sinθ = 2
Rearranging gives:
−2 sin 2 θ + 3 sin θ − 1 = 0
or
2 sin 2 θ − 3 sin θ + 1 = 0
which is a quadratic in sin θ
Using the quadratic formula or by factorising gives:
(2 sinθ − 1)(sin θ − 1) = 0
from which, 2 sinθ − 1 = 0 or sinθ − 1 = 0
and
sin θ =
1
2 or sin θ = 1
from which,
θ = 30
◦ or 150
◦ or 90
◦
Now try the following exercise
Exercise 74 Further problems on double
angles
1. The power p in an electrical circuit is given by
p =
v 2
R
. Determine the power in terms of V ,
R and cos 2t when v = V cos t .
V
2
2R
(1 + cos 2t )
2. Prove the following identities:
(a) 1 −
cos 2φ
cos 2 φ
= tan 2 φ
(b)
1 + cos 2t
sin 2 t
= 2 cot 2 t
(c)
(tan 2x)(1 + tan x)
tan x
=
2
1 − tan x
(d) 2 cosec 2θ cos 2θ = cot θ − tan θ
3. If the third harmonic of a waveform is given by
V 3 cos 3θ, express the third harmonic in terms
of the first harmonic cosθ, when V 3 = 1.
[cos 3θ = 4 cos 3 θ − 3 cosθ]
In Problems 4 to 8, solve for θ in the range
−180 ◦ ≤ θ ≤ 180 ◦
4. cos 2θ = sin θ
[−90 ◦ , 30 ◦ , 150 ◦ ]
5. 3 sin 2θ + 2 cosθ = 0
[−160.47 ◦ , −90 ◦ , −19.47 ◦ , 90 ◦ ]
6. sin 2θ + cos θ = 0
[−150 ◦ , −90 ◦ , −30 ◦ , 90 ◦ ]
7. cos 2θ + 2 sin θ = −3
[ −90 ◦ ]
8. tan θ + cot θ = 2
[ 4 5 ◦ , −135 ◦ ]
17.4 Changing products of sines and
cosines into sums or differences
(i) sin(A + B) + sin(A − B) = 2 sin A cos B (from the
formulae in Section 17.1)
i.e. sin A cos B
=
1
2 [sin(A + B) + sin(A − B)]
(1)
(ii) sin(A + B) − sin(A − B) = 2 cos A sin B
i.e. cos A sin B
=
1
2 [sin(A + B) − sin(A − B)]
(2)
(iii) cos(A + B) + cos(A − B) = 2 cos A cos B
i.e. cos A cos B
=
1
2 [cos(A + B) + cos(A − B)]
(3)