CHAPTER 42 TRIGONOMETRIC IDENTITIES AND EQUATIONS
CHAPTER 42 TRIGONOMETRIC IDENTITIES AND EQUATIONS EXERCISE 175 Page 477
1. Prove the identity: sin x cot x = cos x
1 cos x L.H.S. = sin x cot x = sin x = sin x = cos x = R.H.S. tan x sin x
2. Prove the identity:
L.H.S. =
1
(1 − cos θ )
1 (1 − cos 2 θ )
=
2
=
1 sin θ 2
= cosec θ
(since sin 2 θ + cos 2 θ = 1)
1 = cosec θ = R.H.S. sin θ
3. Prove the identity: 2 cos2 A – 1 = cos2 A – sin2 A R.H.S. = cos 2 A − sin 2= A cos 2 A − (1 − cos 2 A )
since cos 2 A + sin 2 A = 1
2 = cos 2 A − 1 + cos = A 2 cos 2 A − 1 = L.H.S.
4. Prove the identity:
cos x − cos3 x = sin x cos x sin x
cos x − cos3 x cos x(1 − cos 2 x) cos x sin 2 x L.H.S. == = = cos x sin x = sin x cos x = R.H.S. sin x sin x sin x 5. Prove the identity: (1 + cot θ)2 + (1 – cot θ)2 = 2 cosec2θ L.H.S. = (1 + cot θ ) + (1 − cot θ ) = 1 + 2 cot θ + cot 2 θ + 1 − 2 cot θ + cot 2 θ 2
2
= 2 + 2 cot 2 θ = 2 + 2 ( cosec 2 θ − 1) = 2 + 2 cos ec 2θ − 2 = 2 cosec 2 θ = R.H.S. 706
© 2014, John Bird
6. Prove the identity:
sin 2 x(sec x + cosec x) = 1 + tan x cos x tan x
sin 2 x ( sec x + cosec x ) L.H.S. = = cos x tan x
1 1 2 sin x + cos x + sin 2 x sin x cos x sin x cos x sin x = sin x sin x cos x cos x sin x + cos x sin x + cos x = sin x = cos x cos x sin x
=
sin x cos x + = tan x + 1 = 1 + tan x = R.H.S. cos x cos x
707
© 2014, John Bird
EXERCISE 176 Page 479 1. Solve for angles between 0° and 360°: 4 – 7 sin θ = 0
Since 4 – 7 sin θ = 0 then 4 = 7 sin θ and
sin θ =
4 7
4 θ = sin −1 = 34.85° 7
from which,
Since sine is positive, the angle 34.85° occurs in the 1st and 2nd quadrants as shown in the diagram below
Hence, the two angles for θ between 0° and 360° whose sine is 34.85°
and
4 are: 7
180° – 34.85° = 145.15°
2. Solve for angles between 0° and 360°: 3 cosec A + 5.5 = 0
Since 3 cosec A + 5.5 = 0 then 3 cosec A = –5.5
i.e.
from which,
1 5.5 = − sin A 3
or
and
sin A = −
cosec A = −
5.5 3
3 5.5
3 A = sin −1 − = –33.06° 5.5
Since sine is negative, the angle 33.06° occurs in the 3rd and 4th quadrants as shown in the diagram below
708
© 2014, John Bird
Hence, the two angles for A between 0° and 360° whose sine is − 180° + 33.06° = 213.06°
and
3 are: 5.5
360° – 33.06° = 326.94°
3. Solve for angles between 0° and 360°: 4(2.32 – 5.4 cot t) = 0
Since 4(2.32 – 5.4 cot t) = 0 i.e.
Hence,
cot t =
then 2.32 5.4
2.32 – 5.4 cot t = 0 from which,
tan t =
and
2.32 = 5.4 cot t
5.4 2.32
5.4 t = tan −1 = 66.75° 2.32
Since tan is positive, the angle 66.75° occurs in the 1st and 3rd quadrants as shown in the diagram below
Hence, the two angles for t between 0° and 360° whose tan is 66.75°
and
5.4 are: 2.32
180° + 66.75° = 246.75°
4. Solve for θ in the range 0 ≤ θ ≤ 360 : sec θ = 2
Since sec θ = 2
then
cos θ =
1 = 0.5 2
709
© 2014, John Bird
θ = cos −1 0.5 = 60° or 300° since cosine is positive in the 1st and 4th
Hence,
quadrants (check CAST) 5. Solve for θ in the range 0 ≤ θ ≤ 360 : cot θ = 0.6
Since cot θ = 0.6
then
tan θ =
1 = 1.66667 0.6
θ = tan −1 1.66667 = 59° or 239° since tangent is positive in the 1st and 3rd
Hence,
quadrants (check CAST) 6. Solve for θ in the range 0° ≤ θ ≤ 360° : cosec θ = 1.5
Since cosec θ = 1.5 Hence,
then
sin θ =
1 = 0.66667 1.5
θ = sin −1 0.66667 = 41.81° or 138.19° since sine is positive in the 1st and 2nd quadrants (check CAST)
7. Solve for x in the range −180° ≤ x ≤ 180° : sec x = −1.5
Since sec x = −1.5 then Hence,
cos x =
1 = −0.66667 −1.5
θ = cos −1 − 0.66667 = 131.81° or 228.81° since cosine is negative in the 1st and 2nd quadrants (check CAST)
Hence, in the range −180° ≤ x ≤ 180° , θ = 131.81° or – 131.81° i.e.
θ = ±131.81°
8. Solve for x in the range −180° ≤ x ≤ 180° : cot x = 1.2
Since cot x = 1.2
Hence,
then
tan x =
1 1.2
1 x = tan −1 = 39.81° or 210.81° since cosine is positive in the 1st and 1.2
4th quadrants (check CAST)
710
© 2014, John Bird
Hence, in the range −180° ≤ x ≤ 180° , x = 39.81° or – 140.19°
9. Solve for x in the range −180° ≤ x ≤ 180° : cosec x = –2
Since cosec x = –2
Hence,
then
sin x =
1 = − 0.5 −2
x = sin −1 ( − 0.5 ) = 210° or 330° since sine is negative in the 3rd and 4th quadrants (check CAST)
Hence, in the range −180° ≤ x ≤ 180° ,
x = –30° or –150°
10. Solve for θ in the range 0° ≤ θ ≤ 360° : 3sin θ = 2 cos θ
Since
Hence,
3sin θ = 2 cos θ
then
sin θ 2 = cos θ 3
i.e. tan θ =
2
θ = tan −1 = 33.69° or 213.69° 3
2 3
since tangent is positive in the 1st and 3rd quadrants (check CAST)
11. Solve for θ in the range 0° ≤ θ ≤ 360° : 5 cos θ = – sin θ
Since Hence,
5 cos θ = – sin θ
then
sin θ = −5 cos θ
i.e. tan θ = –5
θ = tan −1 (−5) = 101.31° or 281.31°
since tangent is negative in the 2nd and 4th quadrants (check CAST)
711
© 2014, John Bird
EXERCISE 177 Page 479 1. Solve for angles between 0° and 360°: 5 sin2 y = 3
Since 5sin 2 y = 3
then
sin 2 y=
3 = 0.60 5
and
sin y =
0.60 = ± 0.774596...
y = sin −1 (0.774596...) = 50.77°
and
Since sine y is both positive and negative, a value for y occurs in each of the four quadrants, as shown in the diagram below
Hence the values of y between 0° and 360° are: 50.77°,
180° – 50.77° = 129.23°,
180° + 50.77° = 230.77°
and
360° – 50.77° = 309.23°
2. Solve for angles between 0° and 360°: cos 2 θ = 0.25
Since cos 2 θ = 0.25
then
cos θ =
0.25 = ± 0.5
θ = cos −1 (0.5)= 60°
and
Since cos θ is both positive and negative, a value for θ occurs in each of the four quadrants Hence,
θ = 60° or 120° or 240° or 300°
3. Solve for angles between 0° and 360°: tan 2 x = 3
Since and
tan 2 x = 3 then tan x =
3 = ±1.7321
x = tan −1 ( ±1.7321) = 60° or 120° or 240° or 300°
Since tan x is both positive and negative, a value for x occurs in each of the four quadrants 712
© 2014, John Bird
4. Solve for angles between 0° and 360°: 5 + 3 cosec2 D = 8
Since 5 + 3cosec 2 D = then 8 Hence,
1 =1 sin 2 D
3cosec 2 D = 8 − 5 = 3 sin 2 D = 1
and
i.e.
from which,
cosec 2 D = 1
sin D = 1 = ± 1
D = sin −1 ( ± 1)
and
There are two values of D between 0° and 360° which satisfy this equation, as shown in the sinusoidal waveform below
Hence,
D = 90° and 270°
5. Solve for angles between 0° and 360°: 2 cot2 θ = 5
Since 2 cot2 θ = 5
Hence,
1 5 = 2 tan θ 2
from which, and
then
cot 2 θ=
and
5 = 2.5 2
tan 2 θ=
tan θ =
2 = 0.4 5 0.4 = ± 0.63246
θ = tan −1 ( ±0.63246 ) = 32.31° or 147.69° or 212.31° or 327.69°
Since tan θ is both positive and negative, a value for θ occurs in each of the four quadrants
713
© 2014, John Bird
EXERCISE 178 Page 480 1. Solve for angles between 0° and 360°: 15 sin2 A + sin A – 2 = 0
Since 15 sin2 A + sin A – 2 = 0 i.e.
5 sin A + 2 = 0
and
3 sin θ – 1 = 0
then
from which, from which,
0 ( 5sin θ + 2 )( 3sin θ − 1) = sin A = −
2 = − 0.4 5
and= A sin −1 (− 0.4) = –23.58°
1 = 0.3333... 3
sin θ =
and A = sin −1 (0.33333...) =
19.47° From the diagram below, the four values of θ between 0° and 360° are: 19.47°, 180° – 19.47° = 160.53°,
180° + 23.58° = 203.58°
and
360° – 23.58° = 336.42°
2. Solve for angles between 0° and 360°: 8 tan2 θ + 2 tan θ = 15
8 tan 2 θ + 2 tan θ − 15 = 0
Since 8 tan 2 θ + 2 tan θ = 15 then
0 ( 4 tan θ − 5)( 2 tan θ + 3) =
i.e.
5 = 1.25 4
i.e.
4 tan θ – 5 = 0
from which,
tan θ =
and
2 tan θ + 3 = 0
from which,
tan θ = −
3 = −1.5 2
and θ = tan −1 1.25 = 51.34° and= θ tan −1 − 1.5 = –56.31°
From the diagram below, the four values of θ between 0° and 360° are: 51.34°, 180° – 56.31° = 123.69°,
180° + 51.34° = 231.34°
714
and
360° – 56.31° = 303.69°
© 2014, John Bird
3. Solve for angles between 0° and 360°: 2 cosec2 t – 5 cosec t = 12
Since 2 cosec 2 t − 5cosec t = then 12 and
2 cosec 2 t − 5cosec t − 12 = 0 (2 cosec t + 3)(cosec t – 4) = 0
i.e.
2 cosec t + 3 = 0
and
cosec t – 4 = 0
from which, cosec t = −
from which,
3 2
cosec t = 4
sin t = −
and
and
sin t =
2 3
from which, t = –41.81°
1 from which, t = 14.48° 4
From the diagram below, the four values of θ between 0° and 360° are: 14.48°, 180° – 14.48° = 165.52°,
180° + 41.81° = 221.81°
and
360° – 41.81° = 318.19°
4. Solve for angles between 0° and 360°: 2 cos2 θ + 9 cos θ – 5 = 0 Factorizing 2 cos2 θ + 9 cos θ – 5 = 0 gives: (2 cos θ – 1)(cos θ + 5) = 0 from which, either (2 cos θ – 1) = 0 or (cos θ + 5) = 0 1 2
i.e.
cos θ =
or cos θ = –5 (which has no solution)
Hence,
θ = cos −1 = 60° or 300° since cosine is positive in the 1st and 2
1
4th quadrants 715
© 2014, John Bird
EXERCISE 179 Page 481 1. Solve for angles between 0° and 360°: 2 cos 2 θ + sin θ = 1
Since
cos 2 θ + sin 2 θ = 1 then
cos 2 θ = 1 − sin 2 θ
Hence 2 cos 2 θ + sin θ = 1 becomes
2(1 − sin 2 θ ) + sin θ = 1
i.e.
2 − 2sin 2 θ + sin θ = 1
i.e.
2sin 2 θ − sin θ − 1 =0
(2 sin θ + 1)(sin θ – 1) = 0
Factorizing gives: from which, i.e.
Hence,
either (2 sin θ + 1) = 0 sin θ = −
1 2
or (sin θ – 1) = 0 or sin θ = 1
1
θ = sin −1 − = 210° or 330° (since sine is negative in the 3rd and 4th 2 quadrants
or
θ = sin −1 1 = 90°
2. Solve for angles between 0° and 360°: 4 cos 2 t + 5sin t = 3
Since
cos 2 t + sin 2 t = 1 then
cos 2 t = 1 − sin 2 t
Hence 4 cos 2 t + 5sin t = 3 becomes
4(1 − sin 2 t ) + 5sin t = 3
i.e.
4 − 4sin 2 t + 5sin t = 3
i.e.
4sin 2 t − 5sin t − 1 =0
Using the quadratic formula: sin t =
− − 5 ± (−5) 2 − 4(4)(−1) 5 ± 41 = 2(4) 8
= 1.4254… (which has no solution) Hence,
or –0.17539…
t = sin −1 ( −0.17539...) = 191.1° or 349.9° (since sine is negative in the 3rd and 4th quadrants
716
© 2014, John Bird
3. Solve for angles between 0° and 360°: 2 cos θ − 4sin 2 θ = 0
Since Hence,
cos 2 θ + sin 2 θ = 1 then
sin 2 θ = 1 − cos 2 θ
2 cos θ − 4sin 2 θ = 0 becomes
2 cos θ − 4(1 − cos 2 θ ) = 0
i.e.
2 cos θ − 4 + 4 cos 2 θ = 0
i.e.
4 cos 2 θ + 2 cos θ − 4 = 0
or
2 cos 2 θ + cos θ − 2 = 0
−1 ± 12 − 4(2)(−2) −1 ± 17 = 2(2) 4
Using the quadratic = formula: cos θ
= 0.7807764… Hence,
or –1.280776… (which has no solution)
θ = cos −1 0.7807764... = 38.67° or 321.33° since cosine is positive in the 1st and 4th quadrants
4. Solve for angles between 0° and 360°: 3cos θ + 2sin 2 θ = 3
Since Hence,
cos 2 θ + sin 2 θ = 1 then
sin 2 θ = 1 − cos 2 θ
3cos θ + 2(1 − cos 2 θ ) = 3
3cos θ + 2sin 2 θ = 3 becomes
i.e.
3cos θ + 2 − 2 cos 2 θ = 3
i.e.
2 cos 2 θ − 3cos θ + 1 = 0
(2 cos θ – 1)(cos θ – 1) = 0
Factorizing gives: from which, i.e.
either (2 cos θ – 1) = 0 cos θ =
1 2
or
(cos θ – 1) = 0
or cos θ = 1
1
Hence,
θ = cos −1 = 60° or 300° (since cosine is positive in the 1st and 4th quadrants 2
and
θ = cos −1 1 = 0° or 360°
Hence,
θ = 0° or 60° or 300° or 360° 717
© 2014, John Bird
5. Solve for angles between 0° and 360°: 12 sin2 θ – 6 = cos θ
Since 12sin 2 θ − 6 = cos θ
(
)
12 1 − cos 2 θ − 6 = cos θ
then
12 − 12 cos 2 θ − 6 = cos θ
i.e.
12 cos 2 θ + cos θ − 6 = 0
i.e.
(4 cos θ + 3)(3 cos θ – 2) = 0
Factorizing gives: i.e.
4 cos θ + 3 = 0
and
3 cos θ – 2 = 0
from which, from which,
cos θ = − cos θ =
2 3
3 = −0.75 and 4
and
θ = cos −1 (−0.75) = –41.41° 2
θ = cos −1 = 48.19° 3
From the diagram below, the four values of θ between 0° and 360° are: 48.19°, 180° – 41.41° = 138.59°,
180° + 41.41° = 221.41°
and
360° – 48.19° = 311.81°
6. Solve for angles between 0° and 360°: 16 sec x – 2 = 14 tan2 x 1 + tan2 x = sec2 x
from which, tan2 x = sec2 x – 1
Substituting for tan2 x in 16 sec x – 2 = 14 tan2 x gives:
16 sec x – 2 = 14(sec2 x – 1)
i.e.
16 sec x – 2 = 14 sec2 x – 14 14 sec2 x – 16 sec x – 12 = 0
− − 16 ± (−16) 2 − 4(14)(−12) 16 ± 928 Using the quadratic formula: sec x = = 2(14) 28
= 1.659396 or –0.516539 If sec x = 1.659396, then cos x =
1 = 0.602630 1.659396
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© 2014, John Bird
x = cos −1 ( 0.602630 ) = 52.94º or 307.06º, since cosine is positive in the 1st
and
and 4th quadrants or, if sec x = –0.516539 then cos x = Hence,
1 = −1.93596 which is not possible −0.516539
x = 52.94° or 307.06°
7. Solve for angles between 0° and 360°: 4 cot2 A – 6 cosec A + 6 = 0
Since 4 cot 2 A − 6 cosec A + 6 = 0
then
(
)
4 cosec 2 A − 1 − 6 cosec A + 6 = 0 4 cosec 2 A − 6 cosec A + 2 = 0
and Factorizing gives:
(2 cosec A – 1) (2 cosec A – 2) = 0 1 2
and
i.e.
2 cosec A – 1 = 0
from which,
cosec A =
and
2 cosec A – 2 = 0
from which,
cosec A = 1 and
sin A = 2 which has no solutions sin A = 1, which has only one solution
between 0° and 360°, i.e. A = 90°
8. Solve for angles between 0° and 360°: 5 sec t + 2 tan2 t = 3 1 + tan2 x = sec2 x
from which, tan2 x = sec2 x – 1
Substituting for tan2 x in 5 sec t + 2 tan2 t = 3 gives:
5 sec t + 2(sec2 x – 1) = 3
i.e.
5 sec t + 2sec2 x – 2 = 3 2 sec2 x + 5 sec x – 5 = 0 −5 ± (5) 2 − 4(2)(−5) −5 ± 65 = 2(2) 4
Using the quadratic formula: sec t =
= 0.765564 or –3.2655644 If sec t = 0.765564, then cos t =
1 = 1.3062265 which is not possible 0.765564
and sec t = –3.2655644, then cos t =
1 = − 0.30622578 −3.2655644
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© 2014, John Bird
and
t = cos −1 ( − 0.30622578 ) = 107.83° or 252.17°, since cosine is negative in the 2nd and 3rd quadrants
Hence,
t = 107.83° or 252.17°
9. Solve for angles between 0° and 360°: 2.9 cos2 a – 7 sin a + 1 = 0
Since 2.9 cos 2 a − 7 sin a + 1 = 0
2.9(1 − sin 2 a ) − 7 sin a + 1 = 0
then
2.9 − 2.9sin 2 a − 7 sin A + 1 = 0
i.e.
2.9sin 2 a + 7 sin a − 3.9 = 0
and Hence,
Thus,
−7 ± 7 2 − 4(2.9)(−3.9) −7 ± 94.24 sin a = = 0.46685 or –2.88064, which has = 2(2.9) 5.8 no solution a = sin −1 (0.46685) = 27.83°
and, from the diagram below, 180° – 27°50′ = 152.17°
10. Solve for angles between 0° and 360°: 3 cosec2 ß = 8 – 7 cot ß cot 2 β + 1 = cosec 2 β and substituting for cosec 2 β in 3cosec 2 β = 8 − 7 cot β gives:
(
)
3 cot 2 β + 1 =8 − 7 cot β
i.e. and
3cot 2 β + 3 = 8 − 7 cot β 3cot 2 β + 7 cot β − 5 = 0
Using the quadratic formula: −7 ± 7 2 − 4(3)(−5) −7 ± 109 cot β = = = 0.573384 or − 2.906718 2(3) 6 720
© 2014, John Bird
1 from which, = β cot = t −1 0.573384 tan −1 = 60.17° and 240.17° 0.573384
(i.e. 1st and 3rd quadrants) 1 = 161.02° and 341.02° −2.906718
cot −1 (−2.906718) = tan −1 β=
and
(i.e. 2nd and 4th quadrants) 11. Solve for angles between 0° and 360°: cot θ = sin θ
Since cot = θ
1 cos θ = tan θ sin θ
then
cos θ = sin θ sin θ
i.e. cos θ = sin 2 θ
cos 2 θ + sin 2 θ = 1 from which, sin 2 θ = 1 − cos 2 θ
Hence, cos θ = sin 2 θ
becomes:
cos θ = 1 − cos 2 θ
cos 2 θ + cos θ − 1 =0
i.e.
−1 ± 12 − 4(1)(−1) −1 ± 5 = 0.61803398… or – 1.61803398…, which has cos θ = = 2(1) 2 no solution Hence,
θ = cos −1 ( 0.61803398..) = 51.83° and 308.17°
12. Solve for angles between 0° and 360°: tan θ + 3cot θ = 5sec θ
tan θ + 3cot θ = 5sec θ is the same as:
sin θ cos θ 1 + 3 5 = cos θ sin θ cos θ
Multiplying each term by sin θ cos θ gives:
( sin θ cos θ ) Cancelling gives:
sin θ cos θ + 3 ( sin θ cos θ ) cos θ sin θ
1 5 ( sin θ cos θ ) = cos θ
sin 2 θ + 3cos 2 θ = 5sin θ
i.e.
sin 2 θ + 3(1 − sin 2 θ ) = 5sin θ
i.e.
sin 2 θ + 3 − 3sin 2 θ = 5sin θ
and
2sin 2 θ + 5sin θ − 3 = 0
721
© 2014, John Bird
Factorizing gives: from which, i.e.
Hence,
(2 sin θ – 1)(sin θ + 3) = 0 (2 sin θ – 1) = 0 sin θ =
θ = sin −1
1 2
or
or
(sin θ + 3) = 0
sin θ = –3 (which has no solution)
1 = 30° and 150° 2
722
© 2014, John Bird
1. Prove the identity: sin x cot x = cos x
1 cos x L.H.S. = sin x cot x = sin x = sin x = cos x = R.H.S. tan x sin x
2. Prove the identity:
L.H.S. =
1
(1 − cos θ )
1 (1 − cos 2 θ )
=
2
=
1 sin θ 2
= cosec θ
(since sin 2 θ + cos 2 θ = 1)
1 = cosec θ = R.H.S. sin θ
3. Prove the identity: 2 cos2 A – 1 = cos2 A – sin2 A R.H.S. = cos 2 A − sin 2= A cos 2 A − (1 − cos 2 A )
since cos 2 A + sin 2 A = 1
2 = cos 2 A − 1 + cos = A 2 cos 2 A − 1 = L.H.S.
4. Prove the identity:
cos x − cos3 x = sin x cos x sin x
cos x − cos3 x cos x(1 − cos 2 x) cos x sin 2 x L.H.S. == = = cos x sin x = sin x cos x = R.H.S. sin x sin x sin x 5. Prove the identity: (1 + cot θ)2 + (1 – cot θ)2 = 2 cosec2θ L.H.S. = (1 + cot θ ) + (1 − cot θ ) = 1 + 2 cot θ + cot 2 θ + 1 − 2 cot θ + cot 2 θ 2
2
= 2 + 2 cot 2 θ = 2 + 2 ( cosec 2 θ − 1) = 2 + 2 cos ec 2θ − 2 = 2 cosec 2 θ = R.H.S. 706
© 2014, John Bird
6. Prove the identity:
sin 2 x(sec x + cosec x) = 1 + tan x cos x tan x
sin 2 x ( sec x + cosec x ) L.H.S. = = cos x tan x
1 1 2 sin x + cos x + sin 2 x sin x cos x sin x cos x sin x = sin x sin x cos x cos x sin x + cos x sin x + cos x = sin x = cos x cos x sin x
=
sin x cos x + = tan x + 1 = 1 + tan x = R.H.S. cos x cos x
707
© 2014, John Bird
EXERCISE 176 Page 479 1. Solve for angles between 0° and 360°: 4 – 7 sin θ = 0
Since 4 – 7 sin θ = 0 then 4 = 7 sin θ and
sin θ =
4 7
4 θ = sin −1 = 34.85° 7
from which,
Since sine is positive, the angle 34.85° occurs in the 1st and 2nd quadrants as shown in the diagram below
Hence, the two angles for θ between 0° and 360° whose sine is 34.85°
and
4 are: 7
180° – 34.85° = 145.15°
2. Solve for angles between 0° and 360°: 3 cosec A + 5.5 = 0
Since 3 cosec A + 5.5 = 0 then 3 cosec A = –5.5
i.e.
from which,
1 5.5 = − sin A 3
or
and
sin A = −
cosec A = −
5.5 3
3 5.5
3 A = sin −1 − = –33.06° 5.5
Since sine is negative, the angle 33.06° occurs in the 3rd and 4th quadrants as shown in the diagram below
708
© 2014, John Bird
Hence, the two angles for A between 0° and 360° whose sine is − 180° + 33.06° = 213.06°
and
3 are: 5.5
360° – 33.06° = 326.94°
3. Solve for angles between 0° and 360°: 4(2.32 – 5.4 cot t) = 0
Since 4(2.32 – 5.4 cot t) = 0 i.e.
Hence,
cot t =
then 2.32 5.4
2.32 – 5.4 cot t = 0 from which,
tan t =
and
2.32 = 5.4 cot t
5.4 2.32
5.4 t = tan −1 = 66.75° 2.32
Since tan is positive, the angle 66.75° occurs in the 1st and 3rd quadrants as shown in the diagram below
Hence, the two angles for t between 0° and 360° whose tan is 66.75°
and
5.4 are: 2.32
180° + 66.75° = 246.75°
4. Solve for θ in the range 0 ≤ θ ≤ 360 : sec θ = 2
Since sec θ = 2
then
cos θ =
1 = 0.5 2
709
© 2014, John Bird
θ = cos −1 0.5 = 60° or 300° since cosine is positive in the 1st and 4th
Hence,
quadrants (check CAST) 5. Solve for θ in the range 0 ≤ θ ≤ 360 : cot θ = 0.6
Since cot θ = 0.6
then
tan θ =
1 = 1.66667 0.6
θ = tan −1 1.66667 = 59° or 239° since tangent is positive in the 1st and 3rd
Hence,
quadrants (check CAST) 6. Solve for θ in the range 0° ≤ θ ≤ 360° : cosec θ = 1.5
Since cosec θ = 1.5 Hence,
then
sin θ =
1 = 0.66667 1.5
θ = sin −1 0.66667 = 41.81° or 138.19° since sine is positive in the 1st and 2nd quadrants (check CAST)
7. Solve for x in the range −180° ≤ x ≤ 180° : sec x = −1.5
Since sec x = −1.5 then Hence,
cos x =
1 = −0.66667 −1.5
θ = cos −1 − 0.66667 = 131.81° or 228.81° since cosine is negative in the 1st and 2nd quadrants (check CAST)
Hence, in the range −180° ≤ x ≤ 180° , θ = 131.81° or – 131.81° i.e.
θ = ±131.81°
8. Solve for x in the range −180° ≤ x ≤ 180° : cot x = 1.2
Since cot x = 1.2
Hence,
then
tan x =
1 1.2
1 x = tan −1 = 39.81° or 210.81° since cosine is positive in the 1st and 1.2
4th quadrants (check CAST)
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© 2014, John Bird
Hence, in the range −180° ≤ x ≤ 180° , x = 39.81° or – 140.19°
9. Solve for x in the range −180° ≤ x ≤ 180° : cosec x = –2
Since cosec x = –2
Hence,
then
sin x =
1 = − 0.5 −2
x = sin −1 ( − 0.5 ) = 210° or 330° since sine is negative in the 3rd and 4th quadrants (check CAST)
Hence, in the range −180° ≤ x ≤ 180° ,
x = –30° or –150°
10. Solve for θ in the range 0° ≤ θ ≤ 360° : 3sin θ = 2 cos θ
Since
Hence,
3sin θ = 2 cos θ
then
sin θ 2 = cos θ 3
i.e. tan θ =
2
θ = tan −1 = 33.69° or 213.69° 3
2 3
since tangent is positive in the 1st and 3rd quadrants (check CAST)
11. Solve for θ in the range 0° ≤ θ ≤ 360° : 5 cos θ = – sin θ
Since Hence,
5 cos θ = – sin θ
then
sin θ = −5 cos θ
i.e. tan θ = –5
θ = tan −1 (−5) = 101.31° or 281.31°
since tangent is negative in the 2nd and 4th quadrants (check CAST)
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© 2014, John Bird
EXERCISE 177 Page 479 1. Solve for angles between 0° and 360°: 5 sin2 y = 3
Since 5sin 2 y = 3
then
sin 2 y=
3 = 0.60 5
and
sin y =
0.60 = ± 0.774596...
y = sin −1 (0.774596...) = 50.77°
and
Since sine y is both positive and negative, a value for y occurs in each of the four quadrants, as shown in the diagram below
Hence the values of y between 0° and 360° are: 50.77°,
180° – 50.77° = 129.23°,
180° + 50.77° = 230.77°
and
360° – 50.77° = 309.23°
2. Solve for angles between 0° and 360°: cos 2 θ = 0.25
Since cos 2 θ = 0.25
then
cos θ =
0.25 = ± 0.5
θ = cos −1 (0.5)= 60°
and
Since cos θ is both positive and negative, a value for θ occurs in each of the four quadrants Hence,
θ = 60° or 120° or 240° or 300°
3. Solve for angles between 0° and 360°: tan 2 x = 3
Since and
tan 2 x = 3 then tan x =
3 = ±1.7321
x = tan −1 ( ±1.7321) = 60° or 120° or 240° or 300°
Since tan x is both positive and negative, a value for x occurs in each of the four quadrants 712
© 2014, John Bird
4. Solve for angles between 0° and 360°: 5 + 3 cosec2 D = 8
Since 5 + 3cosec 2 D = then 8 Hence,
1 =1 sin 2 D
3cosec 2 D = 8 − 5 = 3 sin 2 D = 1
and
i.e.
from which,
cosec 2 D = 1
sin D = 1 = ± 1
D = sin −1 ( ± 1)
and
There are two values of D between 0° and 360° which satisfy this equation, as shown in the sinusoidal waveform below
Hence,
D = 90° and 270°
5. Solve for angles between 0° and 360°: 2 cot2 θ = 5
Since 2 cot2 θ = 5
Hence,
1 5 = 2 tan θ 2
from which, and
then
cot 2 θ=
and
5 = 2.5 2
tan 2 θ=
tan θ =
2 = 0.4 5 0.4 = ± 0.63246
θ = tan −1 ( ±0.63246 ) = 32.31° or 147.69° or 212.31° or 327.69°
Since tan θ is both positive and negative, a value for θ occurs in each of the four quadrants
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© 2014, John Bird
EXERCISE 178 Page 480 1. Solve for angles between 0° and 360°: 15 sin2 A + sin A – 2 = 0
Since 15 sin2 A + sin A – 2 = 0 i.e.
5 sin A + 2 = 0
and
3 sin θ – 1 = 0
then
from which, from which,
0 ( 5sin θ + 2 )( 3sin θ − 1) = sin A = −
2 = − 0.4 5
and= A sin −1 (− 0.4) = –23.58°
1 = 0.3333... 3
sin θ =
and A = sin −1 (0.33333...) =
19.47° From the diagram below, the four values of θ between 0° and 360° are: 19.47°, 180° – 19.47° = 160.53°,
180° + 23.58° = 203.58°
and
360° – 23.58° = 336.42°
2. Solve for angles between 0° and 360°: 8 tan2 θ + 2 tan θ = 15
8 tan 2 θ + 2 tan θ − 15 = 0
Since 8 tan 2 θ + 2 tan θ = 15 then
0 ( 4 tan θ − 5)( 2 tan θ + 3) =
i.e.
5 = 1.25 4
i.e.
4 tan θ – 5 = 0
from which,
tan θ =
and
2 tan θ + 3 = 0
from which,
tan θ = −
3 = −1.5 2
and θ = tan −1 1.25 = 51.34° and= θ tan −1 − 1.5 = –56.31°
From the diagram below, the four values of θ between 0° and 360° are: 51.34°, 180° – 56.31° = 123.69°,
180° + 51.34° = 231.34°
714
and
360° – 56.31° = 303.69°
© 2014, John Bird
3. Solve for angles between 0° and 360°: 2 cosec2 t – 5 cosec t = 12
Since 2 cosec 2 t − 5cosec t = then 12 and
2 cosec 2 t − 5cosec t − 12 = 0 (2 cosec t + 3)(cosec t – 4) = 0
i.e.
2 cosec t + 3 = 0
and
cosec t – 4 = 0
from which, cosec t = −
from which,
3 2
cosec t = 4
sin t = −
and
and
sin t =
2 3
from which, t = –41.81°
1 from which, t = 14.48° 4
From the diagram below, the four values of θ between 0° and 360° are: 14.48°, 180° – 14.48° = 165.52°,
180° + 41.81° = 221.81°
and
360° – 41.81° = 318.19°
4. Solve for angles between 0° and 360°: 2 cos2 θ + 9 cos θ – 5 = 0 Factorizing 2 cos2 θ + 9 cos θ – 5 = 0 gives: (2 cos θ – 1)(cos θ + 5) = 0 from which, either (2 cos θ – 1) = 0 or (cos θ + 5) = 0 1 2
i.e.
cos θ =
or cos θ = –5 (which has no solution)
Hence,
θ = cos −1 = 60° or 300° since cosine is positive in the 1st and 2
1
4th quadrants 715
© 2014, John Bird
EXERCISE 179 Page 481 1. Solve for angles between 0° and 360°: 2 cos 2 θ + sin θ = 1
Since
cos 2 θ + sin 2 θ = 1 then
cos 2 θ = 1 − sin 2 θ
Hence 2 cos 2 θ + sin θ = 1 becomes
2(1 − sin 2 θ ) + sin θ = 1
i.e.
2 − 2sin 2 θ + sin θ = 1
i.e.
2sin 2 θ − sin θ − 1 =0
(2 sin θ + 1)(sin θ – 1) = 0
Factorizing gives: from which, i.e.
Hence,
either (2 sin θ + 1) = 0 sin θ = −
1 2
or (sin θ – 1) = 0 or sin θ = 1
1
θ = sin −1 − = 210° or 330° (since sine is negative in the 3rd and 4th 2 quadrants
or
θ = sin −1 1 = 90°
2. Solve for angles between 0° and 360°: 4 cos 2 t + 5sin t = 3
Since
cos 2 t + sin 2 t = 1 then
cos 2 t = 1 − sin 2 t
Hence 4 cos 2 t + 5sin t = 3 becomes
4(1 − sin 2 t ) + 5sin t = 3
i.e.
4 − 4sin 2 t + 5sin t = 3
i.e.
4sin 2 t − 5sin t − 1 =0
Using the quadratic formula: sin t =
− − 5 ± (−5) 2 − 4(4)(−1) 5 ± 41 = 2(4) 8
= 1.4254… (which has no solution) Hence,
or –0.17539…
t = sin −1 ( −0.17539...) = 191.1° or 349.9° (since sine is negative in the 3rd and 4th quadrants
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© 2014, John Bird
3. Solve for angles between 0° and 360°: 2 cos θ − 4sin 2 θ = 0
Since Hence,
cos 2 θ + sin 2 θ = 1 then
sin 2 θ = 1 − cos 2 θ
2 cos θ − 4sin 2 θ = 0 becomes
2 cos θ − 4(1 − cos 2 θ ) = 0
i.e.
2 cos θ − 4 + 4 cos 2 θ = 0
i.e.
4 cos 2 θ + 2 cos θ − 4 = 0
or
2 cos 2 θ + cos θ − 2 = 0
−1 ± 12 − 4(2)(−2) −1 ± 17 = 2(2) 4
Using the quadratic = formula: cos θ
= 0.7807764… Hence,
or –1.280776… (which has no solution)
θ = cos −1 0.7807764... = 38.67° or 321.33° since cosine is positive in the 1st and 4th quadrants
4. Solve for angles between 0° and 360°: 3cos θ + 2sin 2 θ = 3
Since Hence,
cos 2 θ + sin 2 θ = 1 then
sin 2 θ = 1 − cos 2 θ
3cos θ + 2(1 − cos 2 θ ) = 3
3cos θ + 2sin 2 θ = 3 becomes
i.e.
3cos θ + 2 − 2 cos 2 θ = 3
i.e.
2 cos 2 θ − 3cos θ + 1 = 0
(2 cos θ – 1)(cos θ – 1) = 0
Factorizing gives: from which, i.e.
either (2 cos θ – 1) = 0 cos θ =
1 2
or
(cos θ – 1) = 0
or cos θ = 1
1
Hence,
θ = cos −1 = 60° or 300° (since cosine is positive in the 1st and 4th quadrants 2
and
θ = cos −1 1 = 0° or 360°
Hence,
θ = 0° or 60° or 300° or 360° 717
© 2014, John Bird
5. Solve for angles between 0° and 360°: 12 sin2 θ – 6 = cos θ
Since 12sin 2 θ − 6 = cos θ
(
)
12 1 − cos 2 θ − 6 = cos θ
then
12 − 12 cos 2 θ − 6 = cos θ
i.e.
12 cos 2 θ + cos θ − 6 = 0
i.e.
(4 cos θ + 3)(3 cos θ – 2) = 0
Factorizing gives: i.e.
4 cos θ + 3 = 0
and
3 cos θ – 2 = 0
from which, from which,
cos θ = − cos θ =
2 3
3 = −0.75 and 4
and
θ = cos −1 (−0.75) = –41.41° 2
θ = cos −1 = 48.19° 3
From the diagram below, the four values of θ between 0° and 360° are: 48.19°, 180° – 41.41° = 138.59°,
180° + 41.41° = 221.41°
and
360° – 48.19° = 311.81°
6. Solve for angles between 0° and 360°: 16 sec x – 2 = 14 tan2 x 1 + tan2 x = sec2 x
from which, tan2 x = sec2 x – 1
Substituting for tan2 x in 16 sec x – 2 = 14 tan2 x gives:
16 sec x – 2 = 14(sec2 x – 1)
i.e.
16 sec x – 2 = 14 sec2 x – 14 14 sec2 x – 16 sec x – 12 = 0
− − 16 ± (−16) 2 − 4(14)(−12) 16 ± 928 Using the quadratic formula: sec x = = 2(14) 28
= 1.659396 or –0.516539 If sec x = 1.659396, then cos x =
1 = 0.602630 1.659396
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© 2014, John Bird
x = cos −1 ( 0.602630 ) = 52.94º or 307.06º, since cosine is positive in the 1st
and
and 4th quadrants or, if sec x = –0.516539 then cos x = Hence,
1 = −1.93596 which is not possible −0.516539
x = 52.94° or 307.06°
7. Solve for angles between 0° and 360°: 4 cot2 A – 6 cosec A + 6 = 0
Since 4 cot 2 A − 6 cosec A + 6 = 0
then
(
)
4 cosec 2 A − 1 − 6 cosec A + 6 = 0 4 cosec 2 A − 6 cosec A + 2 = 0
and Factorizing gives:
(2 cosec A – 1) (2 cosec A – 2) = 0 1 2
and
i.e.
2 cosec A – 1 = 0
from which,
cosec A =
and
2 cosec A – 2 = 0
from which,
cosec A = 1 and
sin A = 2 which has no solutions sin A = 1, which has only one solution
between 0° and 360°, i.e. A = 90°
8. Solve for angles between 0° and 360°: 5 sec t + 2 tan2 t = 3 1 + tan2 x = sec2 x
from which, tan2 x = sec2 x – 1
Substituting for tan2 x in 5 sec t + 2 tan2 t = 3 gives:
5 sec t + 2(sec2 x – 1) = 3
i.e.
5 sec t + 2sec2 x – 2 = 3 2 sec2 x + 5 sec x – 5 = 0 −5 ± (5) 2 − 4(2)(−5) −5 ± 65 = 2(2) 4
Using the quadratic formula: sec t =
= 0.765564 or –3.2655644 If sec t = 0.765564, then cos t =
1 = 1.3062265 which is not possible 0.765564
and sec t = –3.2655644, then cos t =
1 = − 0.30622578 −3.2655644
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© 2014, John Bird
and
t = cos −1 ( − 0.30622578 ) = 107.83° or 252.17°, since cosine is negative in the 2nd and 3rd quadrants
Hence,
t = 107.83° or 252.17°
9. Solve for angles between 0° and 360°: 2.9 cos2 a – 7 sin a + 1 = 0
Since 2.9 cos 2 a − 7 sin a + 1 = 0
2.9(1 − sin 2 a ) − 7 sin a + 1 = 0
then
2.9 − 2.9sin 2 a − 7 sin A + 1 = 0
i.e.
2.9sin 2 a + 7 sin a − 3.9 = 0
and Hence,
Thus,
−7 ± 7 2 − 4(2.9)(−3.9) −7 ± 94.24 sin a = = 0.46685 or –2.88064, which has = 2(2.9) 5.8 no solution a = sin −1 (0.46685) = 27.83°
and, from the diagram below, 180° – 27°50′ = 152.17°
10. Solve for angles between 0° and 360°: 3 cosec2 ß = 8 – 7 cot ß cot 2 β + 1 = cosec 2 β and substituting for cosec 2 β in 3cosec 2 β = 8 − 7 cot β gives:
(
)
3 cot 2 β + 1 =8 − 7 cot β
i.e. and
3cot 2 β + 3 = 8 − 7 cot β 3cot 2 β + 7 cot β − 5 = 0
Using the quadratic formula: −7 ± 7 2 − 4(3)(−5) −7 ± 109 cot β = = = 0.573384 or − 2.906718 2(3) 6 720
© 2014, John Bird
1 from which, = β cot = t −1 0.573384 tan −1 = 60.17° and 240.17° 0.573384
(i.e. 1st and 3rd quadrants) 1 = 161.02° and 341.02° −2.906718
cot −1 (−2.906718) = tan −1 β=
and
(i.e. 2nd and 4th quadrants) 11. Solve for angles between 0° and 360°: cot θ = sin θ
Since cot = θ
1 cos θ = tan θ sin θ
then
cos θ = sin θ sin θ
i.e. cos θ = sin 2 θ
cos 2 θ + sin 2 θ = 1 from which, sin 2 θ = 1 − cos 2 θ
Hence, cos θ = sin 2 θ
becomes:
cos θ = 1 − cos 2 θ
cos 2 θ + cos θ − 1 =0
i.e.
−1 ± 12 − 4(1)(−1) −1 ± 5 = 0.61803398… or – 1.61803398…, which has cos θ = = 2(1) 2 no solution Hence,
θ = cos −1 ( 0.61803398..) = 51.83° and 308.17°
12. Solve for angles between 0° and 360°: tan θ + 3cot θ = 5sec θ
tan θ + 3cot θ = 5sec θ is the same as:
sin θ cos θ 1 + 3 5 = cos θ sin θ cos θ
Multiplying each term by sin θ cos θ gives:
( sin θ cos θ ) Cancelling gives:
sin θ cos θ + 3 ( sin θ cos θ ) cos θ sin θ
1 5 ( sin θ cos θ ) = cos θ
sin 2 θ + 3cos 2 θ = 5sin θ
i.e.
sin 2 θ + 3(1 − sin 2 θ ) = 5sin θ
i.e.
sin 2 θ + 3 − 3sin 2 θ = 5sin θ
and
2sin 2 θ + 5sin θ − 3 = 0
721
© 2014, John Bird
Factorizing gives: from which, i.e.
Hence,
(2 sin θ – 1)(sin θ + 3) = 0 (2 sin θ – 1) = 0 sin θ =
θ = sin −1
1 2
or
or
(sin θ + 3) = 0
sin θ = –3 (which has no solution)
1 = 30° and 150° 2
722
© 2014, John Bird
