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ESCUELA SUPERIOR POLITÉCNICA DE CHIMBORAZO 
 
 
 
 
 
 
 
 
 
 
 
 
PRIMER SEMESTRE 
PARALELO ¨B¨ 
ANÁLISIS MATEMÁTICO I 
 
BORIS JOSUE ASQUI VACA 
2020-2021
 𝑟 = 1 − cos 𝜃 ; 𝜗 =
𝜋
3
 
• 𝐶𝑜𝑟𝑑𝑖𝑜𝑖𝑑𝑒 
0 <
1
1
< 1 
• 𝑉𝑒𝑟𝑖𝑓𝑖𝑐𝑎𝑟 𝑠𝑖 𝑟 = 𝑓(𝜃) 𝑝𝑎𝑠𝑎 𝑝𝑜𝑟 𝑒𝑙 𝑝𝑜𝑙𝑜 
𝑟 = 0 
1 − cos 𝜃 = 0 
𝜃 = cos−1(1) 
𝜃 = 0 
• 𝐼𝑛𝑡𝑒𝑟𝑠𝑒𝑐𝑐𝑖ó𝑛 𝑒𝑗𝑒 𝑝𝑜𝑙𝑎𝑟 (0, 𝜋); 𝑒𝑗𝑒 𝑝𝑜𝑙𝑎𝑟 90𝑜 (
𝜋
2
,
3𝜋
2
) 
𝜃 𝑟 (𝑟, 𝜃) 
0 0 (0,0) 
𝜋
2
 1 (1,
𝜋
2
) 
𝜋 2 (2, 𝜋) 
3𝜋
2
 
1 
(1,
3𝜋
2
) 
4)𝐷𝑎𝑡𝑜𝑠 
𝜋
6
 
𝜋
3
 
2𝜋
3
 
5𝜋
6
 
7𝜋
6
 
4𝜋
3
 
5𝜋
3
 
11𝜋
6
 
0.1339 0.5 1.5 1.8660 1.8660 1.5 0.5 0.1339 
 
• 𝐻𝑎𝑙𝑙𝑎𝑟( 𝜇; 𝜑) → 𝜃 =
𝜋
3
= 60𝑜 
tan 𝜇 =
1 − cos 𝜃
sin 𝜃
=
1 − cos(60𝑜)
sin 60𝑜
=
√3
3
 
𝜇 = arctan
√3
3
→ 30𝑜 = 𝜇 =
𝜋
6
 
𝜑 = 𝜃 + 𝜇 
𝜑 = 60 + 30 
𝜑 = 90 
tan 𝜑 =
𝑟′ sin 𝜃 + 𝑟 cos 𝜃 
𝑟′ cos 𝜃 − 𝑟 sin 𝜃
 
 
𝑟 = 1 − cos 𝜃 
𝑟′ = sin 𝜃 
tan 𝜑 =
(sin 𝜃) sin 𝜃 + (1 − cos 𝜃) cos 𝜃 
(sin 𝜃) cos 𝜃 − (1 − cos 𝜃) sin 𝜃
 
tan 𝜑 =
sin2 𝜃 + cos 𝜃 − cos2 𝜃 
2(sin 𝜃) cos 𝜃 − sin 𝜃
 
• 𝑇𝑎𝑛𝑔𝑒𝑛𝑡𝑒𝑠 
𝐻𝑜𝑟𝑖𝑧𝑜𝑛𝑡𝑎𝑙𝑒𝑠 
sin2 𝜃 + cos 𝜃 − cos2 𝜃 = 0 
1 − cos2 𝜃 + cos 𝜃 − cos2 𝜃 = 0 
−2 cos2 𝜃 + cos 𝜃 + 1 = 0 
𝑉𝑒𝑟𝑡𝑖𝑐𝑎𝑙𝑒𝑠 
2(sin 𝜃) cos 𝜃 − sin 𝜃 = 0 
sin 𝜃 (2 cos 𝜃 − 1) = 0 
cos 𝜃 = 1 
𝜃 = 0 
𝜃 = 0 + 2𝜋𝑘 
 
𝑘 = 0 
𝜃 = 0 + 2𝜋(0) 
𝜃 = 0 
 
𝑘 = 1 
𝜃 = 0 + 2𝜋(1) 
𝜃 = 2𝜋 
 
𝑘 = −1 
𝜃 = 0 + 2𝜋(−1) 
𝜃 = 0 − 2𝜋 
𝜃 = −2𝜋 
cos 𝜃 = −
1
2
 
𝜃 =
2𝜋
3
 
𝜃 = (−1)𝑘
2𝜋
3
+ 𝜋𝑘 
 
𝑘 = 0 
𝜃 = (−1)0
2𝜋
3
+ 𝜋𝑘 
𝜃 =
2𝜋
3
 
 
𝑘 = 1 
𝜃 = (−1)1
𝜋
3
+ 𝜋(1) 
𝜃 =
2
3
𝜋 
 
𝑘 = −1 
𝜃 = (−1)−1
𝜋
3
+ 𝜋(−1) 
𝜃 = −
4
3
𝜋 
sin 𝜃 = 0 
𝜃 = 0 
𝜃 = 0 + 𝜋𝑘 
 
𝑘 = 0 
𝜃 = 0 + 𝜋(0) 
𝜃 = 0 
 
𝑘 = 1 
𝜃 = 0 + 𝜋(1) 
𝜃 = 𝜋 
 
𝑘 = −1 
𝜃 = 0 + 𝜋(−1) 
𝜃 = −𝜋 
 
 
cos 𝜃 =
1
2
 
𝜃 = 𝜋/3 
𝜃 =
𝜋
3
+ 2𝜋𝑘 
 
𝑘 = 0 
𝜃 =
𝜋
3
+ 2𝜋(0) 
𝜃 =
𝜋
3
 
 
𝑘 = 1 
𝜃 =
𝜋
3
+ 2𝜋(1) 
𝜃 =
7
3
𝜋 
 
𝑘 = −1 
𝜃 =
𝜋
3
+ 2𝜋(−1) 
𝜃 =
𝜋
3
 
 
 𝑟 = 2 − 3sin 𝜃 ; 𝜗 =
𝜋
4
 
• 𝐿𝑖𝑚𝑎𝑧𝑜𝑛 
0 <
2
3
< 1 
• 𝑉𝑒𝑟𝑖𝑓𝑖𝑐𝑎𝑟 𝑠𝑖 𝑟 = 𝑓(𝜃) 𝑝𝑎𝑠𝑎 𝑝𝑜𝑟 𝑒𝑙 𝑝𝑜𝑙𝑜 
𝑟 = 0 
2 − 3sin 𝜃 = 0 
𝜃 = sin−1 (
2
3
) 
𝜃 = 41.81 
• 𝐼𝑛𝑡𝑒𝑟𝑠𝑒𝑐𝑐𝑖ó𝑛 𝑒𝑗𝑒 𝑝𝑜𝑙𝑎𝑟 (0, 𝜋); 𝑒𝑗𝑒 𝑝𝑜𝑙𝑎𝑟 90𝑜 (
𝜋
2
,
3𝜋
2
) 
𝜃 𝑟 (𝑟, 𝜃) 
0 2 (2,0) 
𝜋
2
 −1 (−1,
𝜋
2
) 
𝜋 2 (2, 𝜋) 
3𝜋
2
 
5 
(5,
3𝜋
2
) 
4)𝐷𝑎𝑡𝑜𝑠 
𝜋
6
 
𝜋
3
 
2𝜋
3
 
5𝜋
6
 
7𝜋
6
 
4𝜋
3
 
5𝜋
3
 
11𝜋
6
 
0.5 0.5 -0.60 0.5 3.5 4.60 4.60 3.5 
 
• 𝐻𝑎𝑙𝑙𝑎𝑟 (𝜇; 𝜑) → 𝜃 =
𝜋
4
= 45𝑜 
tan 𝜇 =
2 − 3sin 𝜃
−3 cos 𝜃
=
2 − 3sin(45)
−3 cos(45)
= 0.057 
𝜇 = arctan 0.057 → 𝜇 = 3.27𝑜 
𝜑 = 𝜃 + 𝜇 
𝜑 = 45 + 3.27 
𝜑 = 48.27 
tan 𝜑 =
𝑟′ sin 𝜃 + 𝑟 cos 𝜃 
𝑟′ cos 𝜃 − 𝑟 sin 𝜃
 
 
𝑟 = 2 − 3sin 𝜃 
𝑟′ = −3 cos 𝜃 
tan 𝜑 =
(−3 cos 𝜃) sin 𝜃 + (2 − 3sin 𝜃) cos 𝜃 
(−3 cos 𝜃) cos 𝜃 − (2 − 3sin 𝜃) sin 𝜃
 
tan 𝜑 =
−6 cos 𝜃 𝑠𝑖𝑛𝜃 + 2 cos 𝜃
−3 cos2 𝜃 − 2 sin 𝜃 + 3 sin2 𝜃
 
• 𝑇𝑎𝑛𝑔𝑒𝑛𝑡𝑒𝑠 
𝐻𝑜𝑟𝑖𝑧𝑜𝑛𝑡𝑎𝑙𝑒𝑠 
−6 cos 𝜃 𝑠𝑖𝑛𝜃 + 2 cos 𝜃 = 0 
2 cos 𝜃 (−3 sin 𝜃 + 1) = 0 
𝑉𝑒𝑟𝑡𝑖𝑐𝑎𝑙𝑒𝑠 
−3(1 − sin2 𝜃) − 2 sin 𝜃 + 3 sin2 𝜃 
6 sin2 𝜃 − 2 sin 𝜃 − 3 = 0 
2 cos 𝜃 = 0 
𝜃 =
𝜋
2
 
𝜃 =
𝜋
2
+ 𝜋𝑘 
 
𝑘 = 0 
𝜃 =
𝜋
2
+ 𝜋(0) 
𝜃 =
𝜋
2
 
 
𝑘 = 1 
𝜃 =
𝜋
2
+ 𝜋(1) 
𝜃 =
3𝜋
2
 
 
𝑘 = −1 
𝜃 =
𝜋
2
+ 𝜋(−1) 
𝜃 = −
𝜋
2
 
 
sin 𝜃 =
1
3
 
𝜃 = 0.1081𝜋 
𝜃 = (−1)𝑘0.1081𝜋 + 𝜋𝑘 
 
𝑘 = 0 
𝜃 = (−1)00.1081𝜋 + 𝜋(0) 
𝜃 = 0.1081𝜋 
 
𝑘 = 1 
𝜃 = (−1)10.1081𝜋 + 𝜋(1) 
𝜃 = 0.8919𝜋 
 
𝑘 = −1 
𝜃 = (−1)−10.1081𝜋 + 𝜋(−1) 
𝜃 = −1.1081𝜋 
sin 𝜃 = 0.8931 
𝜃 = 0.3515𝜋 
𝜃 = 0.3515𝜋 + 𝜋𝑘 
 
𝑘 = 0 
𝜃 = 0.3515𝜋 + 𝜋(0) 
𝜃 = 0.3515𝜋 
 
𝑘 = 1 
𝜃 = 0.3515𝜋 + 𝜋(1) 
𝜃 = 1.3515𝜋 
 
𝑘 = −1 
𝜃 = 0.3515𝜋 + 𝜋(−1) 
𝜃 = −0.6485𝜋 
 
 
cos 𝜃 = −0.5598 
𝜃 =
65
36
𝜋 
𝜃 =
65
36
𝜋 + 2𝜋𝑘 
 
𝑘 = 0 
𝜃 =
65
36
𝜋 + 2𝜋(0) 
𝜃 =
65
36
𝜋 
𝑘 = 1 
𝜃 =
65
36
𝜋 + 2𝜋(1) 
𝜃 =
137
36
𝜋 
 
𝑘 = −1 
𝜃 =
65
36
𝜋 + 2𝜋(−1) 
𝜃 = −
7
36
𝜋 
 
 𝑟 = 3 + 2 sin 𝜃 ; 𝜗 =
𝜋
6
 
• 
0 <
3
2
< 1 
• 𝑉𝑒𝑟𝑖𝑓𝑖𝑐𝑎𝑟 𝑠𝑖 𝑟 = 𝑓(𝜃) 𝑝𝑎𝑠𝑎 𝑝𝑜𝑟 𝑒𝑙 𝑝𝑜𝑙𝑜 
𝑟 = 0 
3 + 2 sin 𝜃 = 0 
𝜃 = sin−1 (−
3
2
) 
𝜃 = 𝑆𝑖𝑛 𝑠𝑜𝑙𝑢𝑐𝑖ó𝑛 
• 𝐼𝑛𝑡𝑒𝑟𝑠𝑒𝑐𝑐𝑖ó𝑛 𝑒𝑗𝑒 𝑝𝑜𝑙𝑎𝑟 (0, 𝜋); 𝑒𝑗𝑒 𝑝𝑜𝑙𝑎𝑟 90𝑜 (
𝜋
2
,
3𝜋
2
) 
𝜃 𝑟 (𝑟, 𝜃) 
0 3 (3,0) 
𝜋
2
 5 (5,
𝜋
2
) 
𝜋 3 (3, 𝜋) 
3𝜋
2
 
1 
(1,
3𝜋
2
) 
4)𝐷𝑎𝑡𝑜𝑠 
𝜋
6
 
𝜋
3
 
2𝜋
3
 
5𝜋
6
 
7𝜋
6
 
4𝜋
3
 
5𝜋
3
 
11𝜋
6
 
4 4.73 4.73 4 2 1.26 1.26 2 
 
• 𝐻𝑎𝑙𝑙𝑎𝑟 (𝜇; 𝜑) → 𝜃 =
𝜋
6
= 30𝑜 
tan 𝜇 =
3 + 2 sin 𝜃
2 cos 𝜃
=
3 + 2 sin(30𝑜)
2 cos(30𝑜)
= 2.3094 
𝜇 = arctan 2.3094 → 𝜇 = 65.58𝑜 
𝜑 = 𝜃 + 𝜇 
𝜑 = 30 + 65.58 
𝜑 = 96.58 
tan 𝜑 =
𝑟′ sin 𝜃 + 𝑟 cos 𝜃 
𝑟′ cos 𝜃 − 𝑟 sin 𝜃
 
 
𝑟 = 3 + 2 sin 𝜃 
𝑟′ = 2 cos 𝜃 
tan 𝜑 =
(2 cos 𝜃) sin 𝜃 + (3 + 2 sin 𝜃) cos 𝜃 
(2 cos 𝜃) cos 𝜃 − (3 + 2 sin 𝜃) sin 𝜃
 
tan 𝜑 =
4 sin 𝜃 cos 𝜃 + 3 cos 𝜃
2 cos2 𝜃 − 3 sin 𝜃 − 2 sin2 𝜃
 
• 𝑇𝑎𝑛𝑔𝑒𝑛𝑡𝑒𝑠 
𝐻𝑜𝑟𝑖𝑧𝑜𝑛𝑡𝑎𝑙𝑒𝑠 
4 sin 𝜃 cos 𝜃 + 3 cos 𝜃 = 0 
cos 𝜃 (4 sin 𝜃 + 3) = 0 
𝑉𝑒𝑟𝑡𝑖𝑐𝑎𝑙𝑒𝑠 
2(1 − sin2 𝜃) − 3 sin 𝜃 − 2 sin2 𝜃 
−4 sin2 𝜃 − 3 sin 𝜃 + 2 = 0 
cos 𝜃 = 0 
𝜃 =
𝜋
2
 
𝜃 =
𝜋
2
+ 𝜋𝑘 
 
𝑘 = 0 
𝜃 =
𝜋
2
+ 𝜋(0) 
𝜃 =
𝜋
2
 
 
𝑘 = 1 
𝜃 =
𝜋
2
+ 𝜋(1) 
𝜃 =
3𝜋
2
 
 
𝑘 = −1 
𝜃 =
𝜋
2
+ 𝜋(−1) 
𝜃 = −
𝜋
2
 
 
sin 𝜃 =
−3
4
 
𝜃 = 1.73𝜋 
𝜃 = (−1)𝑘1.73𝜋 + 𝜋𝑘 
 
𝑘 = 0 
𝜃 = (−1)01.73𝜋 + 𝜋(0) 
𝜃 = 1.73𝜋 
 
𝑘 = 1 
𝜃 = (−1)11.73𝜋 + 𝜋(1) 
𝜃 = −0.73𝜋 
 
𝑘 = −1 
𝜃 = (−1)−11.73𝜋 + 𝜋(−1) 
𝜃 = −2.73𝜋 
sin 𝜃 = 1.1753 
 
 
sin 𝜃 = 0.4253 
𝜃 =
5
36
𝜋 
𝜃 =
5
36
𝜋 + 2𝜋𝑘 
 
𝑘 = 0 
𝜃 =
5
36
𝜋 + 2𝜋(0) 
𝜃 =
5
36
𝜋 
 
𝑘 = 1 
𝜃 =
5
36
𝜋 + 2𝜋(1) 
𝜃 =
77
36
𝜋 
 
𝑘 = −1 
𝜃 =
5
36
𝜋 + 2𝜋(−1) 
𝜃 = −
67
36
𝜋 
 
 
 𝑟 = 3 + 2 cos 𝜃 ; 𝜗 =
𝜋
9
 
0 <
3
2
< 1 
• 𝑉𝑒𝑟𝑖𝑓𝑖𝑐𝑎𝑟 𝑠𝑖 𝑟 = 𝑓(𝜃) 𝑝𝑎𝑠𝑎 𝑝𝑜𝑟 𝑒𝑙 𝑝𝑜𝑙𝑜 
𝑟 = 0 
3 + 2 cos 𝜃 = 0 
𝜃 = cos−1 (−
3
2
) 
𝜃 = 𝑁𝑜 ℎ𝑎𝑦 𝑠𝑜𝑙𝑢𝑐𝑖ó𝑛 
• 𝐼𝑛𝑡𝑒𝑟𝑠𝑒𝑐𝑐𝑖ó𝑛 𝑒𝑗𝑒 𝑝𝑜𝑙𝑎𝑟 (0, 𝜋); 𝑒𝑗𝑒 𝑝𝑜𝑙𝑎𝑟 90𝑜 (
𝜋
2
,
3𝜋
2
) 
𝜃 𝑟 (𝑟, 𝜃) 
0 5 (5,0) 
𝜋
2
 3 (3,
𝜋
2
) 
𝜋 1 (1, 𝜋) 
3𝜋
2
 
3 
(3,
3𝜋
2
) 
4)𝐷𝑎𝑡𝑜𝑠 
𝜋
6
 
𝜋
3
 
2𝜋
3
 
5𝜋
6
 
7𝜋
6
 
4𝜋
3
 
5𝜋
3
 
11𝜋
6
 
4.7320 4 2 1.2679 1.2679 2 4 4.7320 
 
• 𝐻𝑎𝑙𝑙𝑎𝑟( 𝜇; 𝜑) → 𝜃 =
𝜋
9
= 20𝑜 
tan 𝜇 =
3 + 2 cos 𝜃
−2 sin 𝜃
=
3 + 2 cos(20)
−2 sin(20)
= −7.1331 
𝜇 = arctan −7.1331 → 278𝑜 = 𝜇 =
139
90
𝜋 
𝜑 = 𝜃 + 𝜇 
𝜑 = 20 + 278 
𝜑 = 298𝑜 
tan 𝜑 =
𝑟′ sin 𝜃 + 𝑟 cos 𝜃 
𝑟′ cos 𝜃 − 𝑟 sin 𝜃
 
 
 
𝑟 = 3 + 2 cos 𝜃 
𝑟′ = −2 sin 𝜃 
tan 𝜑 =
(−2 sin 𝜃) sin 𝜃 + (3 + 2 cos 𝜃) cos 𝜃 
(−2 sin 𝜃) cos 𝜃 − (3 + 2 cos 𝜃) sin 𝜃
 
tan 𝜑 =
−2 sin2 𝜃 + 3 cos 𝜃 + 2 cos2 𝜃 
−2 sin 𝜃 cos 𝜃 − 3 sin 𝜃 − 2 cos 𝜃 sin 𝜃
 
• 𝑇𝑎𝑛𝑔𝑒𝑛𝑡𝑒𝑠 
𝐻𝑜𝑟𝑖𝑧𝑜𝑛𝑡𝑎𝑙𝑒𝑠 
−2 sin2 𝜃 + 3 cos 𝜃 + 2 cos2 𝜃 = 0 
−2 (1 − cos2 𝜃) + 3 cos 𝜃 + 2 cos2 𝜃 = 0 
4 cos2 𝜃 + 3 cos 𝜃 − 2 = 0 
𝑉𝑒𝑟𝑡𝑖𝑐𝑎𝑙𝑒𝑠 
−2 sin 𝜃 cos 𝜃 − 3 sin 𝜃 − 2 cos 𝜃 sin 𝜃 
sin 𝜃 (−4 cos 𝜃 − 3) = 0 
cos 𝜃 = 0.4253 
𝜃 =
13
36
𝜋 
𝜃 =
13
36
𝜋 + 2𝜋𝑘 
 
𝑘 = 0 
𝜃 =
13
36
𝜋 + 2𝜋(0) 
𝜃 =
13
36
𝜋 
 
𝑘 = 1 
𝜃 =
13
36
𝜋 + 2𝜋(1) 
𝜃 =
85
36
𝜋 
 
𝑘 = −1 
𝜃 =
13
36
𝜋 + 2𝜋(−1) 
𝜃 =
59
36
𝜋 
cos 𝜃 = −1.1753 
 
sin 𝜃 = 0 
𝜃 = 0 
𝜃 = 0 + 𝜋𝑘 
 
𝑘 = 0 
𝜃 = 0 + 𝜋(0) 
𝜃 = 0 
 
𝑘 = 1 
𝜃 = 0 + 𝜋(1) 
𝜃 = 𝜋 
 
𝑘 = −1 
𝜃 = 0 + 𝜋(−1) 
𝜃 = −𝜋 
 
 
cos 𝜃 = −
3
4
 
𝜃 =
23
30
𝜋 
𝜃 =
23
30
𝜋 + 2𝜋𝑘 
 
𝑘 = 0 
𝜃 =
23
30
𝜋 + 2𝜋(0) 
𝜃 =
23
30
𝜋 
 
𝑘 = 1 
𝜃 =
23
30
𝜋 + 2𝜋(1) 
𝜃 =
83
30
𝜋 
 
𝑘 = −1 
𝜃 =
23
30
𝜋 + 2𝜋(−1) 
𝜃 = −
37
30
𝜋 
 
 𝑟 = 1 − sin2 𝜃; 𝜗 =
𝜋
6
 
• 𝐿𝑒𝑚𝑛𝑖𝑠𝑐𝑎𝑡𝑎𝑠 
• 𝑉𝑒𝑟𝑖𝑓𝑖𝑐𝑎𝑟 𝑠𝑖 𝑟 = 𝑓(𝜃) 𝑝𝑎𝑠𝑎 𝑝𝑜𝑟 𝑒𝑙 𝑝𝑜𝑙𝑜 
𝑟 = 0 
cos2 𝜃 = 0 
𝜃 = (cos−1(1))2 
𝜃 = 1 
• 𝐼𝑛𝑡𝑒𝑟𝑠𝑒𝑐𝑐𝑖ó𝑛 𝑒𝑗𝑒 𝑝𝑜𝑙𝑎𝑟 (0, 𝜋); 𝑒𝑗𝑒 𝑝𝑜𝑙𝑎𝑟 90𝑜 (
𝜋
2
,
3𝜋
2
) 
𝜃 𝑟 (𝑟, 𝜃) 
0 1 (1,0) 
𝜋
2
 0 (0,
𝜋
2
) 
𝜋 1 (1, 𝜋) 
3𝜋
2
 
0 
(0,
3𝜋
2
) 
4)𝐷𝑎𝑡𝑜𝑠 
𝜋
6
 
𝜋
3
 
2𝜋
3
 
5𝜋
6
 
7𝜋
6
 
4𝜋
3
 
5𝜋
3
 
11𝜋
6
 
3/4 1/4 1.4 3/4 3/4 1.4 1/4 3/4 
 
• 𝐻𝑎𝑙𝑙𝑎𝑟( 𝜇; 𝜑) → 𝜃 =
𝜋
6
= 30𝑜 
tan 𝜇 =
1 − sin2 𝜃
−sin 2𝜃
=
1 − sin2(30)
−sin(60)
=
−√3
2
 
𝜇 = arctan
−√3
2
→ 320𝑜 = 𝜇 =
16𝜋
9
 
𝜑 = 𝜃 + 𝜇 
𝜑 = 320𝑜 + 30𝑜 
𝜑 = 350𝑜 
tan 𝜑 =
𝑟′ sin 𝜃 + 𝑟 cos 𝜃 
𝑟′ cos 𝜃 − 𝑟 sin 𝜃
 
 
 
 
𝑟 = 1 − sin2 𝜃 
𝑟′ = −sin 2𝜃 
tan 𝜑 =
(−2 cos 𝜃 sin 𝜃) sin 𝜃 + (cos2 𝜃) cos 𝜃 
(−2 cos 𝜃 sin 𝜃) cos 𝜃 − (cos2 𝜃) sin 𝜃
 
tan 𝜑 =
−2 cos 𝜃 (1 − cos2 𝜃) + cos3 𝜃 
−3(1 − sin2 𝜃) sin 𝜃
 
• 𝑇𝑎𝑛𝑔𝑒𝑛𝑡𝑒𝑠 
𝐻𝑜𝑟𝑖𝑧𝑜𝑛𝑡𝑎𝑙𝑒𝑠 
−2 cos 𝜃 (1 − cos2 𝜃) + cos3 𝜃 = 0 
3 cos3 𝜃 − 2 cos 𝜃 = 0 
cos 𝜃 (3 cos2 𝜃 − 2) = 0 
𝑉𝑒𝑟𝑡𝑖𝑐𝑎𝑙𝑒𝑠 
−3(1 − sin2 𝜃) sin 𝜃 = 0 
−3sin 𝜃 (1 − sin2 𝜃) = 0 
cos 𝜃 = 0 
𝜃 =
𝜋
2
 
𝜃 =
𝜋
2
+ 2𝜋𝑘 
 
𝑘 = 0 
𝜃 =
𝜋
2
+ 2𝜋(0) 
𝜃 =
𝜋
2
 
 
𝑘 = 1 
𝜃 =
𝜋
2
+ 2𝜋(1) 
𝜃 =
5
2
𝜋 
 
𝑘 = −1 
𝜃 =
𝜋
2
+ 2𝜋(−1) 
𝜃 =
𝜋
2
 
cos2 𝜃 =
2
3
 
𝜃 =
7𝜋
36
 
𝜃 = (−1)𝑘
7𝜋
36
+ 𝜋𝑘 
 
𝑘 = 0 
𝜃 = (−1)0
7𝜋
36
+ 𝜋𝑘 
𝜃 =
7𝜋
36
 
 
𝑘 = 1 
𝜃 = (−1)1
7𝜋
36
+ 𝜋(1) 
𝜃 =
29
36
𝜋 
 
𝑘 = −1 
𝜃 = (−1)−1
7𝜋
36
+ 𝜋(−1) 
𝜃 =
29
36
𝜋 
−3 sin 𝜃 = 0 
𝜃 = 0 
𝜃 = 0 + 𝜋𝑘 
 
𝑘 = 0 
𝜃 = 0 + 𝜋(0) 
𝜃 = 0 
 
𝑘 = 1 
𝜃 = 0 + 𝜋(1) 
𝜃 = 𝜋 
 
𝑘 = −1 
𝜃 = 0 + 𝜋(−1) 
𝜃 = −𝜋 
 
 
cos2 𝜃 = 0 
𝜃 =
𝜋
2
 
𝜃 =
𝜋
2
+ 2𝜋𝑘 
 
𝑘 = 0 
𝜃 =
𝜋
2
+ 2𝜋(0) 
𝜃 =
𝜋
2
 
 
𝑘 = 1 
𝜃 =
𝜋
2
+ 2𝜋(1) 
𝜃 =
5
2
𝜋 
 
𝑘 = −1 
𝜃 =
𝜋
2
+ 2𝜋(−1) 
𝜃 =
𝜋
2