University of Leeds
School of Mathematics
CIVE 1619
Preparatory mathematics for engineers
Solutions 3
1. (a) First look for solutions in the range 0◦ 6 θ 6 360◦ . If (x, y) is
√ on the unit circle at
2
2
2
angle θ, then x = 1/2. Now x + y = 1 gives y = 3/4, so y = ± 3/2. Thus (x, y) can
be one of the
√ following:
3/2), at angle θ = cos−1 (1/2) = 60◦ , or
(1/2, √
(1/2, − 3/2), at angle 360◦ − 60◦ = 300◦ .
Solutions in the range 360◦ 6 θ 6 720◦ can be found by adding 360◦ to those already
found, giving 360◦ + 60◦ = 420◦ and 360◦ + 300◦ = 660◦ . In all: 60◦ , 300◦ , 420◦ , 660◦ .
(b) First look for solutions in the range 0◦ 6 θ 6 360◦ . If (x, y) is on the unit circle at
angle θ, then y = −0.4. Now x2 + y 2 = 1 gives x2 = 0.84, so x = ±0.917. Thus (x, y)
can be one of the following:
(0.917, −0.4), at angle θ = 360◦ − sin−1 (0.4) = 336.4◦ , or
(−0.917, −0.4), at angle 180◦ + sin−1 (0.4) = 203.6◦ .
Now neither of these is in the required range, but subtracting 360◦ from each gives the
required answer: −23.6◦ and −156.4◦ .
(c) If (x, y) is on the unit circle at angle θ, then y/x = −1,
√ so y = −x. Substituting in
x2 + y 2 = 1 gives x2 + (−x)2 = 1, so 2x2 = 1, so x = ±1/ 2. Thus (x, y) can be one of
the following:
√
√
(1/ √
2, −1/√2), at angle θ = 3π/4, or
(−1/ 2, 1/ 2), at angle 2π − π/4 = 7π/4.
Thus θ = 3π/4 or 7π/4. (In decimal form, 2.356 or 5.498 radians.)
2. (a) Write sec θ, cosec θ as fractions and put them over a common denominator. Then
use sin2 θ + cos2 θ = 1:
1
1
sin2 θ + cos2 θ
1
+
=
=
= sec2 θ cosec2 θ.
2
2
cos2 θ sin θ
cos2 θ sin θ
cos2 θ sin2 θ
(b) (sin θ + cos θ)2 = sin2 θ + cos2 θ + 2 sin θ cos θ = 1 + 2 sin θ cos θ.
Similarly, (sin θ − cos θ)2 = sin2 θ + cos2 θ − 2 sin θ cos θ = 1 − 2 sin θ cos θ.
Adding these together gives
sec2 θ + cosec2 θ =
(sin θ + cos θ)2 + (sin θ − cos θ)2 = (1 + 2 sin θ cos θ) + (1 − 2 sin θ cos θ) = 2.
(c) By the addition formula,
cos(θ + 45◦ ) = cos θ cos 45◦ − sin θ sin 45◦ =
√1
2
cos θ −
√1
2
sin θ =
√1 (cos θ
2
− sin θ).
Thus
cos2 (θ + 45◦ ) = 12 (cos θ − sin θ)2 = 12 (cos2 θ + sin2 θ − 2 sin θ cos θ)
= 12 (1 − 2 sin θ cos θ) = 12 − sin θ cos θ.
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(d) By the double angle formula, cos 2θ = 2 cos2 θ − 1. Thus
1 + cos 2θ
1 + (2 cos2 θ − 1)
2 cos2 θ
cos2 θ
=
=
=
.
1 − cos 2θ
1 − (2 cos2 θ − 1)
2 − 2 cos2 θ
1 − cos2 θ
Using 1 − cos2 θ = sin2 θ, we can write this as cos2 θ/ sin2 θ = cot2 θ.
3. We have
sin(75◦ ) = sin(30◦ + 45◦ ) = sin 30◦ cos 45◦ + cos 30◦ sin 45◦
=
1 √1
2 2
√
+
3 √1
2
2
=
√
1+√ 3
2 2
√
=
√
2+ 6
4
=
√
1+√ 3
,
2 2
and
cos(75◦ ) = cos(30◦ + 45◦ ) = cos 30◦ cos 45◦ − sin 30◦ sin 45◦
√
=
3 √1
2
2
−
1 √1
2 2
√
=
3−1
√
2 2
√
=
√
6− 2
4
√
=
3−1
√ .
2 2
√
√
√
3+1
(1 + 3)/2 2
√ = √
. If you want, you can simplify this by
Thus tan(75 ) = √
( 3 − 1)/2 √
2
3−1
multiplying top and bottom by 3 + 1, giving
√
√
√
√
( 3 + 1)( 3 + 1)
3+2 3+1
◦
√
tan(75 ) = √
=
= 2 + 3.
3−1
( 3 − 1)( 3 + 1)
◦
4. First, sin 3θ = sin(θ + 2θ) = sin θ cos 2θ + cos θ sin 2θ. Now sin 2θ = 2 sin θ cos θ and
cos 2θ = 1 − 2 sin2 θ, so
sin 3θ = sin θ(1 − 2 sin2 θ) + cos θ(2 sin θ cos θ) = sin θ − 2 sin3 θ + 2 sin θ cos2 θ.
Now use cos2 θ = 1 − sin2 θ to write this as
sin θ − 2 sin3 θ + 2 sin θ(1 − sin2 θ) = sin θ − 2 sin3 θ + 2 sin θ − 2 sin3 θ = 3 sin θ − 4 sin3 θ.
√
√
2 + 33 =
10, and α√ is the angle for the point on the unit
5. (a) We have
r
=
1
√
√
circle (3/ 10, 1/ 10). Thus α = sin−1 (1/ 10). In decimal form this is r = 3.16
and α = 18.4◦ .
q
√
√
√
(b) r = (−8)2 + 102 = 164, and α is the angle for (10/ 164, −8/ 164). Thus
√
α = − sin−1 (8/ 164). In decimal form this is r = 12.8 and α = −38.7◦ (or equivalently
α = 321.3◦ ).
(c) r =
√
22 + 0 = 2 and α is the angle for (0, 1). Thus α = 90◦ .
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6. (a) 6x5 .
(b) 30x14 .
(c) The function can be written as 14 x = 14 x1 , so its derivative is 14 x0 = 14 .
(d) 52 x3/2 .
(e) √
− 52 x−3/2 .
(f) 3 x = x1/3 , so its derivative is 13 x−2/3 .
(g) 2x + 3.
(h) 3x2 + 8x.
(i) 6x − 5x−2 + 12 x−1/2 .
7. (a) The expression is of the form y = uv, where u = x5 and v = cos x. Then
dv
and dx
= − sin x, so the product rule gives
du
dx
= 5x4
dy
du
dv
=
v+u
= 5x4 cos x − x5 sin x.
dx
dx
dx
(b) Apply the product rule with u = cos x and v = sin x. Then
dy
dv
= cos x, so dx
= − sin2 x + cos2 x.
dx
du
dx
= − sin x and
√
(c) Apply the product
rule
with
u
=
x and v = cos x + sin x. Then u = x1/2 , so
√
dv
du
1 −1/2
= 2x
= 1/(2 x), and dx = − sin x + cos x, giving
dx
√
1
dy
= √ (cos x + sin x) + x(cos x − sin x).
dx
2 x
(d) The expression is of the form y = u/v where u = x2 + 1 and v = cos x. Then
dv
and dx
= − sin x, so the quotient rule gives
dy
=
dx
du
v
dx
du
dx
= 2x
dv
− u dx
2x cos x + (x2 + 1) sin x
=
.
v2
cos2 x
(e) Apply the quotient rule with u = x + 1 and v = x2 + x + 1. Then
dv
= 2x + 1, so
dx
du
dx
= 1 and
dy
(x2 + x + 1) − (x + 1)(2x + 1)
x2 + 2x
=
=
−
.
dx
(x2 + x + 1)2
(x2 + x + 1)2
(f) Apply the quotient rule with u = sin x and v = sin x + 1. Then
dv
= cos x, so
dx
cos x(sin x + 1) − sin x cos x
cos x
dy
=
.
=
2
dx
(sin x + 1)
(sin x + 1)2
du
dx
= cos x and
(g) sec x = 1/ cos x, so apply the quotient rule with u = 1 and v = cos x. Then
dv
and dx
= − sin x, so
dy
0 − 1 × (− sin x)
sin x
=
=
.
2
dx
cos x
cos2 x
This can be written as sec x tan x if desired.
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du
dx
=0
(h) cot x = cos x/ sin x, so apply the quotient rule with u = cos x and v = sin x. Then
du
dv
= − sin x and dx
= cos x, so
dx
− sin2 x − cos2 x
1
dy
=
= − 2 = − cosec2 x.
2
dx
sin x
sin x
8. In all of these we use the chain rule.
(a) We write y = cos u where u = 3x. Then
dy
du
du
dx
= − sin u and
= 3, so
dy du
dy
=
= (− sin u) × 3 = −3 sin 3x.
dx
du dx
(b) We write y = sin u where u = x2 . Then
dy
du
= cos u and
du
dx
= 2x, so
dy
= (cos u) × 2x = 2x cos x2 .
dx
(c) We write y = sin u where u = x2 + x + 1. Then
dy
du
= cos u and
du
dx
= 2x + 1, so
dy
= (cos u) × (2x + 1) = (2x + 1) cos(x2 + x + 1).
dx
√
dy
(d) We write y = u = u1/2 where u = 1 − x2 . Then du
= 12 u−1/2 = 2√1 u and
so
1
x
dy
= √ (−2x) = − √
.
dx
2 u
1 − x2
(e) We write y = u5 where u = cos x. Then
dy
du
= 5u4 and
du
dx
= − sin x, so
dy
= 5u4 (− sin x) = −5 sin x cos4 x.
dx
(f) We write y = u10 where u = 3x2 + 1. Then
dy
du
= 10u9 and
dy
= 10u9 × 6x = 60x(3x2 + 1)9 .
dx
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du
dx
= 6x, so
du
dx
= −2x,