Trigonometric waveforms 137
angles between 0
◦ and 360
◦ whose tangent is 1.7629
are 60 ◦ 26 and 180 ◦ + 60 ◦ 26 , i.e. 240 ◦ 26 .
Problem 3. Solve sec −1 (−2.1499) =α for angles
of α between 0 ◦ and 360 ◦ .
Secant is negative in the second and third quadrants (i.e. the same as for cosine). From Fig. 14.8,
θ = sec −1 2.1499 =cos −1
1
2.1499
= 62 ◦ 17 .
Measured from 0 ◦ , the two angles between 0 ◦ and 360 ◦
whose secant is −2.1499 are
α = 180
◦
− 62
◦ 17
= 117
◦ 43
and
α = 180
◦
+ 62
◦ 17
= 242
◦ 17
S
1808
2708
08
3608
908
T
A
C
Figure 14.8
Problem 4. Solve cot −1 1.3111 =α for angles of
α between 0 ◦ and 360 ◦ .
Cotangent is positive in the first and third quadrants (i.e. same as for tangent). From Fig. 14.9,
θ = cot −1 1.3111 = tan −1
1
1.3111
= 37 ◦ 20 .
S
1808
2708
08
3608
908
T
A
C
Figure 14.9
Hence
α = 37
◦ 20
and
α = 180
◦
+ 37
◦ 20
= 217
◦ 20
Now try the following exercise
Exercise 61 Further problems on
evaluating trigonometric ratios of any
magnitude
1. Find all the angles between 0
◦ and 360
◦ whose
sine is −0.7321.
[227 ◦ 4 and 312 ◦ 56 ]
2. Determine the angles between 0 ◦ and 360 ◦
whose cosecant is 2.5317.
[23 ◦ 16 and 156 ◦ 44 ]
3. If cotangent x =−0.6312, determine the values of x in the range 0 ◦ ≤ x≤ 360 ◦ .
[122 ◦ 16 and 302 ◦ 16 ]
In Problems 4 to 6 solve the given equations.
4. cos −1 (−0.5316) =t
[t = 122 ◦ 7 and 237 ◦ 53 ]
5. sec −1 2.3162 = x
[x = 64 ◦ 25 and 295 ◦ 35 ]
6. tan −1 0.8314 = θ
[θ = 39 ◦ 44 and 219 ◦ 44 ]
14.3 The production of a sine and
cosine wave
In Fig. 14.10, let OR be a vector 1 unit long and
free to rotate anticlockwise about O. In one revolution a circle is produced and is shown with
15 ◦ sectors. Each radius arm has a vertical and
a horizontal component. For example, at 30 ◦ , the
vertical component is T S and the horizontal component
is OS.
From trigonometric ratios,
sin 30
◦
=
TS
TO
=
TS
1
, i.e. TS = sin 30
◦
and cos 30
◦
=
OS
TO
=
OS
1
, i.e. OS = cos 30
◦
The vertical component TS may be projected across
to T S , which is the corresponding value of 30 ◦
on the graph of y against angle x ◦ . If all such
vertical components as TS are projected on to the
angles between 0
◦ and 360
◦ whose tangent is 1.7629
are 60 ◦ 26 and 180 ◦ + 60 ◦ 26 , i.e. 240 ◦ 26 .
Problem 3. Solve sec −1 (−2.1499) =α for angles
of α between 0 ◦ and 360 ◦ .
Secant is negative in the second and third quadrants (i.e. the same as for cosine). From Fig. 14.8,
θ = sec −1 2.1499 =cos −1
1
2.1499
= 62 ◦ 17 .
Measured from 0 ◦ , the two angles between 0 ◦ and 360 ◦
whose secant is −2.1499 are
α = 180
◦
− 62
◦ 17
= 117
◦ 43
and
α = 180
◦
+ 62
◦ 17
= 242
◦ 17
S
1808
2708
08
3608
908
T
A
C
Figure 14.8
Problem 4. Solve cot −1 1.3111 =α for angles of
α between 0 ◦ and 360 ◦ .
Cotangent is positive in the first and third quadrants (i.e. same as for tangent). From Fig. 14.9,
θ = cot −1 1.3111 = tan −1
1
1.3111
= 37 ◦ 20 .
S
1808
2708
08
3608
908
T
A
C
Figure 14.9
Hence
α = 37
◦ 20
and
α = 180
◦
+ 37
◦ 20
= 217
◦ 20
Now try the following exercise
Exercise 61 Further problems on
evaluating trigonometric ratios of any
magnitude
1. Find all the angles between 0
◦ and 360
◦ whose
sine is −0.7321.
[227 ◦ 4 and 312 ◦ 56 ]
2. Determine the angles between 0 ◦ and 360 ◦
whose cosecant is 2.5317.
[23 ◦ 16 and 156 ◦ 44 ]
3. If cotangent x =−0.6312, determine the values of x in the range 0 ◦ ≤ x≤ 360 ◦ .
[122 ◦ 16 and 302 ◦ 16 ]
In Problems 4 to 6 solve the given equations.
4. cos −1 (−0.5316) =t
[t = 122 ◦ 7 and 237 ◦ 53 ]
5. sec −1 2.3162 = x
[x = 64 ◦ 25 and 295 ◦ 35 ]
6. tan −1 0.8314 = θ
[θ = 39 ◦ 44 and 219 ◦ 44 ]
14.3 The production of a sine and
cosine wave
In Fig. 14.10, let OR be a vector 1 unit long and
free to rotate anticlockwise about O. In one revolution a circle is produced and is shown with
15 ◦ sectors. Each radius arm has a vertical and
a horizontal component. For example, at 30 ◦ , the
vertical component is T S and the horizontal component
is OS.
From trigonometric ratios,
sin 30
◦
=
TS
TO
=
TS
1
, i.e. TS = sin 30
◦
and cos 30
◦
=
OS
TO
=
OS
1
, i.e. OS = cos 30
◦
The vertical component TS may be projected across
to T S , which is the corresponding value of 30 ◦
on the graph of y against angle x ◦ . If all such
vertical components as TS are projected on to the