168 Higher Engineering Mathematics
i.e.
θ + 59.03
◦
= 43.32
◦ or 136.68
◦
Hence θ = 43.32
◦
− 59.03
◦
= −15.71
◦
or
θ = 136.68
◦
− 59.03
◦
= 77.65
◦
Since −15.71 ◦ is the same as −15.71 ◦ + 360 ◦ , i.e.
344.29 ◦ , then the solutions are θ = 77.65
◦ or 344.29
◦ ,
which may be checked by substituting into the original
equation.
Problem 10. Solve the equation
3.5 cos A − 5.8 sin A = 6.5 for 0 ◦ ≤ A ≤ 360 ◦ .
Let 3.5 cos A − 5.8 sin A = R sin(A + α)
= R[sin A cos α + cos A sin α]
= (R cos α) sin A + (R sin α) cos A
Equating coefficients gives:
3.5 = R sin α, from which, sin α =
3.5
R
and −5.8 = R cos α, from which, cosα =
−5.8
R
There is only one quadrant in which both sine is positive and cosine is negative, i.e. the second, as shown in
Fig. 17.7.
2708
908
3608
1808
08
␣
R
25.8
3.5
Figure 17.7
From Fig. 17.7, R =
[(3.5) 2 + (−5.8) 2 ]= 6.774 and
θ = tan
−1 3.5
5.8
= 31.12
◦ .
Hence α = 180 ◦ − 31.12 ◦ = 148.88 ◦ .
Thus
3.5 cos A − 5.8 sin A = 6.774 sin(A + 144.88
◦
) = 6.5
Hence sin(A + 148.88
◦
) =
6.5
6.774
, from which,
(A + 148.88
◦
) = sin
−1 6.5
6.774
= 73.65
◦ or 106.35
◦
Thus A = 73.65
◦
− 148.88
◦
= −75.23
◦
≡ (−75.23
◦
+ 360
◦
) = 284.77
◦
or
A = 106.35
◦
− 148.88
◦
= −42.53
◦
≡ (−42.53
◦
+ 360
◦
) = 317.47
◦
The solutions are thus A = 284.77 ◦ or 317.47 ◦ , which
may be checked in the original equation.
Now try the following exercise
Exercise 73 Further problems on the
conversion of a sin ω t + b cos ω t into
R sin(ω t + α)
In Problems 1 to 4, change the functions into the
form R sin(ωt ± α).
1. 5 sin ωt + 8 cosωt [9.434 sin(ωt + 1.012)]
2. 4 sin ωt − 3 cosωt
[5 sin(ωt − 0.644)]
3. −7 sinωt + 4 cos ωt
[8.062 sin(ωt + 2.622)]
4. −3 sinωt − 6 cos ωt
[6.708 sin(ωt − 2.034)]
5. Solve the following equations for values of θ
between 0
◦ and 360
◦ : (a) 2 sinθ + 4 cos θ = 3
(b) 12 sin θ − 9 cosθ = 7.
(a) 74.44 ◦ or 338.70 ◦
(b) 64.69 ◦ or 189.05 ◦
6. Solve the following equations for
0 ◦ < A < 360 ◦ : (a) 3cosA + 2 sin A = 2.8
(b) 12 cos A − 4 sin A = 11.
(a) 72.73 ◦ or 354.63 ◦
(b) 11.15 ◦ or 311.98 ◦
7. Solve the following equations for values of θ
between 0 ◦ and 360 ◦ : (a) 3 sinθ + 4 cosθ = 3
(b) 2 cosθ + sin θ = 2.
[(a) 90 ◦ or 343.74 ◦ (b) 0 ◦ , 53.14 ◦ ]
i.e.
θ + 59.03
◦
= 43.32
◦ or 136.68
◦
Hence θ = 43.32
◦
− 59.03
◦
= −15.71
◦
or
θ = 136.68
◦
− 59.03
◦
= 77.65
◦
Since −15.71 ◦ is the same as −15.71 ◦ + 360 ◦ , i.e.
344.29 ◦ , then the solutions are θ = 77.65
◦ or 344.29
◦ ,
which may be checked by substituting into the original
equation.
Problem 10. Solve the equation
3.5 cos A − 5.8 sin A = 6.5 for 0 ◦ ≤ A ≤ 360 ◦ .
Let 3.5 cos A − 5.8 sin A = R sin(A + α)
= R[sin A cos α + cos A sin α]
= (R cos α) sin A + (R sin α) cos A
Equating coefficients gives:
3.5 = R sin α, from which, sin α =
3.5
R
and −5.8 = R cos α, from which, cosα =
−5.8
R
There is only one quadrant in which both sine is positive and cosine is negative, i.e. the second, as shown in
Fig. 17.7.
2708
908
3608
1808
08
␣
R
25.8
3.5
Figure 17.7
From Fig. 17.7, R =
[(3.5) 2 + (−5.8) 2 ]= 6.774 and
θ = tan
−1 3.5
5.8
= 31.12
◦ .
Hence α = 180 ◦ − 31.12 ◦ = 148.88 ◦ .
Thus
3.5 cos A − 5.8 sin A = 6.774 sin(A + 144.88
◦
) = 6.5
Hence sin(A + 148.88
◦
) =
6.5
6.774
, from which,
(A + 148.88
◦
) = sin
−1 6.5
6.774
= 73.65
◦ or 106.35
◦
Thus A = 73.65
◦
− 148.88
◦
= −75.23
◦
≡ (−75.23
◦
+ 360
◦
) = 284.77
◦
or
A = 106.35
◦
− 148.88
◦
= −42.53
◦
≡ (−42.53
◦
+ 360
◦
) = 317.47
◦
The solutions are thus A = 284.77 ◦ or 317.47 ◦ , which
may be checked in the original equation.
Now try the following exercise
Exercise 73 Further problems on the
conversion of a sin ω t + b cos ω t into
R sin(ω t + α)
In Problems 1 to 4, change the functions into the
form R sin(ωt ± α).
1. 5 sin ωt + 8 cosωt [9.434 sin(ωt + 1.012)]
2. 4 sin ωt − 3 cosωt
[5 sin(ωt − 0.644)]
3. −7 sinωt + 4 cos ωt
[8.062 sin(ωt + 2.622)]
4. −3 sinωt − 6 cos ωt
[6.708 sin(ωt − 2.034)]
5. Solve the following equations for values of θ
between 0
◦ and 360
◦ : (a) 2 sinθ + 4 cos θ = 3
(b) 12 sin θ − 9 cosθ = 7.
(a) 74.44 ◦ or 338.70 ◦
(b) 64.69 ◦ or 189.05 ◦
6. Solve the following equations for
0 ◦ < A < 360 ◦ : (a) 3cosA + 2 sin A = 2.8
(b) 12 cos A − 4 sin A = 11.
(a) 72.73 ◦ or 354.63 ◦
(b) 11.15 ◦ or 311.98 ◦
7. Solve the following equations for values of θ
between 0 ◦ and 360 ◦ : (a) 3 sinθ + 4 cosθ = 3
(b) 2 cosθ + sin θ = 2.
[(a) 90 ◦ or 343.74 ◦ (b) 0 ◦ , 53.14 ◦ ]