University of Leeds
School of Civil Engineering
CIVE1619, Autumn 2010
Solutions 3
Trigonometry
1.
a) By using the definitions of tan and cot, and writing the fractions over a common denominator gives
sin θ
cos θ
sin2 θ + cos2 θ
tan θ + cot θ =
+
=
.
cos θ
sin θ
cos θ sin θ
Now sin2 θ + cos2 θ = 1, so
tan θ + cot θ =
1
= sec θ cosec θ.
cos θ sin θ
b) By expanding the squares we get
(sin θ + cos θ)2 = sin2 θ + cos2 θ + 2 sin θ cos θ = 1 + 2 sin θ cos θ
and
(sin θ − cos θ)2 = sin2 θ + cos2 θ − 2 sin θ cos θ = 1 − 2 sin θ cos θ.
Adding these together gives
(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
1
1
cos(θ + 45◦ ) = cos(θ) cos(45◦ ) − sin(θ) sin(45◦ ) = √ cos(θ) − √ sin(θ)
2
2
1
= √ (cos θ − sin θ).
2
Thus
cos2 (θ + 45◦ ) =
=
1
1
(cos θ − sin θ)2 = (cos2 θ + sin2 θ − 2 sin θ cos θ)
2
2
1
1
(1 − 2 sin θ cos θ) = − sin θ cos θ.
2
2
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
=
=
=
2
2
1 − cos 2θ
1 − (2 cos θ − 1)
2 − 2 cos θ
1 − cos2 θ
Using that 1 − cos2 θ = sin2 θ this becomes cos2 θ/ sin2 θ = cot2 θ.
e) We have
1 − tan2 θ 1 −
=
1 + tan2 θ 1 +
=
2.
sin2 θ
cos2 θ
sin2 θ
cos2 θ
=
cos2 θ − sin2 θ
cos2 θ + sin2 θ
cos2 θ − sin2 θ
= cos2 θ − sin2 θ = cos(2θ).
1
a) The general solution is θ = cos−1 (1/2) + n360◦ = 60◦ + n360 and θ = − cos−1 (1/2) +
n360◦ = −60◦ + n360◦ . Among the first set of solutions, only 60◦ is in the specified range,
and among the second set of solutions, only −60◦ + 360◦ = 300◦ is in the specified range.
The answer is therefore 60◦ and 300◦ .
1
b) The general solution is θ = sin−1 (−0.4) + n360◦ = −23.6◦ + n360◦ and θ = 180◦ −
sin−1 (−0.4) + n360◦ = 203.6◦ + n360◦ . Among the first set of solutions, only −23.6◦ is in
the specified range, and among the second set of solutions, only 203.6◦ − 360◦ = −156.4◦
is in the specified range. The answer is therefore −23.6◦ and −156.4◦ .
c) The general solution is θ = tan−1 (−1) + nπ = −π/4 + nπ. There are two solutions in
the specified range, −π/4 + π = 3π/4 and −π/4 + 2π = 7π/4, and this is therefore the
answer.
3. We have
√
√
1 3 1 1
3−1
√ ,
sin(15 ) = sin(45 − 30 ) = sin 45 cos 30 − sin 30 cos 45 = √
− √ =
2
2
2
2
2 2
and
√
√
1 3
1 1
3+1
◦
◦
◦
◦
◦
◦
◦
√ .
cos(15 ) = cos(45 − 30 ) = cos 45 cos 30 + sin 45 sin 30 = √
+√
=
2 2
22
2 2
Thus
√
√
√
( 3 − 1)/2 2
3−1
◦
√ =√
tan(15 ) = √
.
( 3 + 1)/2 2
3+1
√
If you want you can simplify this by multiplying top and bottom by 3 − 1, giving
√
√
√
√
( 3 − 1)( 3 − 1)
3−2 3+1
◦
√
tan(15 ) = √
= 2 − 3.
=
3−1
( 3 + 1)( 3 − 1)
◦
4.
◦
◦
◦
◦
◦
◦
√
√
a) We√have r√= 12 + 33 = 10, and
√ α is the angle for the point on the unit circle
(1/ 10, 3/ 10). Thus α = sin−1 (3/ 10). In decimals this is r = 3.16 and α = 71.6◦ .
p
√
√
√
√
b) r = (−4)2 + 52 = 41, and α is the angle for (5/ 41, −4/ 41). Thus α = sin−1 (−4/ 41).
In decimals this is r = 6.40 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◦ .
5. By using the suggested hint we get
cos(3θ) = cos(θ + 2θ) = cos θ cos(2θ) − sin θ sin(2θ).
Now sin(2θ) = 2 sin θ cos θ and cos(2θ) = 2 cos2 θ − 1, so
cos(3θ) = cos θ(2 cos2 θ − 1) − sin θ(2 sin θ cos θ)
=2 cos3 θ − cos θ − 2 cos θ sin2 θ
=2 cos3 θ − cos θ − 2 cos θ(1 − cos2 θ)
=4 cos3 θ − 3 cos θ.
2
Differentiation
6.
b) 32x15 ;
c)
1
;
4
7 5/2
x ;
2
5
e) − x−3/2 ;
2
f)
1 −3/4
x
;
4
g) 2x + 3;
h) 3x2 + 8x;
1
i) 6x − 5x−2 + x−1/2 .
2
a) 7x6 ;
d)
7. In all of these we use the chain rule.
dy
du
a) We write y = cos u where u = 3x. Now
= − sin u and
= 3, so
du
dx
dy du
dy
=
= (− sin u) × 3 = −3 sin 3x
dx
du dx
b) We write y = sin u where u = x2 . Now
2x cos x2 .
dy
du
dy
= cos u and
= 2x, so
= (cos u) × 2x =
du
dx
dx
du
dy
= cos u and
= 2x + 1, so
c) We write y = sin u where u = x2 + x + 1. Now
du
dx
dy
= (cos u) × (2x + 1) = (2x + 1) cos(x2 + x + 1).
dx
d) We write y =
√
u = u1/2 where u = 1 − x2 .
dy
1
1
du
= u−1/2 = √ and
= −2x, so
du
2
dx
2 u
dy
1
x
= √ (−2x) = − √
dx
2 u
1 − x2
e) We write y = u5 where u = cos x. Now Now
5u4 (− sin x) = −5 sin x cos4 x.
f ) We write y = u10 where u = 3x2 + 1. Now
60x(3x2 + 1)9 .
8.
du
dy
dy
= 5u4 and
= − sin x, so
=
du
dx
dx
dy
du
dy
= 10u9 and
= 6x, so
= 10u9 × 6x =
du
dx
dx
du
a) The expression is of the form y = uv where u = x6 and v = sin x. Now
= 6x5 and
dx
dv
= cos x, so the product rule gives
dx
dy
du
dv
=
v+u
= 6x5 sin x + x6 cos x.
dx
dx
dx
b) Apply the product rule with u = cos x and v = sin x. Now
so
dy
dx
= − sin2 x + cos2 x.
c) Apply the product rule with u =
and
dv
dx
√
3
du
= − sin x and
dx
x = x1/3 and v = cos x + sin x. Now
= − sin x + cos x, so
dy
x−2/3 (cos x + sin x) √
=
+ 3 x(cos x − sin x).
dx
3
3
dv
dx
= cos x,
du
1
= x−2/3
dx
3
d) The expression is of the form y = u/v where u = x2 + 1 and v = cos x. Now
dv
dx
du
= 2x and
dx
= − sin x, so the quotient rule gives
dy
=
dx
du
dx v
dv
− u dx
2x cos x + (x2 + 1) sin x
=
.
v2
cos2 x
du
dv
e) Apply the quotient rule with u = x + 1 and v = x2 + x + 1. Now
= 1 and
= 2x + 1,
dx
dx
so
(x2 + x + 1) − (x + 1)(2x + 1)
dy
=
dx
(x2 + x + 1)2
x2 + 2x
=− 2
.
(x + x + 1)2
du
dv
f ) Apply the quotient rule with u = sin x and v = sin x + 1. Now
= cos x and
= cos x,
dx
dx
so
dy
cos x(sin x + 1) − sin x cos x
=
dx
(sin x + 1)2
cos x
=
.
(sin x + 1)2
du
g) sec x = 1/ cos x, so apply the quotient rule with u = 1 and v = cos x. Now
= 0 and
dx
dv
= − sin x, so
dx
0 − 1 × (− sin x)
sin x
dy
=
=
.
dx
cos2 x
cos2 x
This can be written as sec x tan x if desired.
h) cot x = cos x/ sin x, so apply the quotient rule with u = cos x and v = sin x. Now
du
dv
dx = − sin x and dx = cos x, so
− sin2 x − cos2 x
1
dy
=
= − 2 = − cosec2 x.
2
dx
sin x
sin x
du
= 0 and
i) cosec x = 1/ sin x, so apply the quotient rule with u = 1 and v = sin x. Now
dx
dv
= cos x, so
dx
dy
0 − 1 × (cos x)
cos x
=
=− 2 .
2
dx
sin x
sin x
This can be written as − cosec x cot x if desired.
4