Find the all values in the given table.

Question

Want to see more full solutions like this?

Answer 1

Question

Chapter E, Problem 1PS

To determine

Find the all values in the given table.

Expert Solution & Answer

Answer to Problem 1PS

The complete table is

Degree	Radians	Sine	Cosine	Tangent
−45° 180° −240° 210° 90°	−π4 π−4π3 7π6 π2	−12 032 −12 1	12 −1−12 −32 0	−10−3 13 undefined

Explanation of Solution

Given:

The given table is

Student's Solution and Survival Manual for Calculus, Chapter E, Problem 1PS

Calculation:

180°=πc1°=πc180−45°=πc180×−45°=−πc4sin(− 45°)=−sin( 45°)=−12cos(− 45°)=cos( 45°)=12tan(− 45°)=−tan( 45°)=−1

πc=180°sin( 180°)=0cos( 180°)=−1tan( 180°)=0

πc=180°πc= 180°π−4π3= 180°π×(− 4π3)=−240°sin(− 240°)=−sin( 240°)=−sin( 180°+ 60°)=sin( 60°)=32cos(− 240°)=cos( 240°)=cos( 180°+ 60°)=−cos( 60°)=−12tan(− 240°)=−tan( 240°)=−tan( 180°+ 60°)=−tan( 60°)=−3

180°=πc1°=πc180210°=πc180×210°=( 7π 6)csin( 210°)=sin( 180°+ 30°)=−sin( 30°)=−12cos( 210°)=cos( 180°+ 30°)=−cos( 30°)=−32tan( 210°)=tan( 180°+ 30°)=tan( 30°)=13

180°=πc1°=πc18090°=πc180×90°=( π 2)csin( 90°)=1cos( 90°)=0tan( 90°)=undefined

Hence the complete table is

Degree	Radians	Sine	Cosine	Tangent
−45° 180° −240° 210° 90°	−π4 π−4π3 7π6 π2	−12 032 −12 1	12 −1−12 −32 0	−10−3 13 undefined

Want to see more full solutions like this?

Subscribe now to access step-by-step solutions to millions of textbook problems written by subject matter experts!

Students have asked these similar questions

3.1 Limits 1. If lim f(x)=-6 and lim f(x)=5, then lim f(x). Explain your choice. x+3° x+3* x+3 (a) Is 5 (c) Does not exist (b) is 6 (d) is infinite

1 pts Let F and G be vector fields such that ▼ × F(0, 0, 0) = (0.76, -9.78, 3.29), G(0, 0, 0) = (−3.99, 6.15, 2.94), and G is irrotational. Then sin(5V (F × G)) at (0, 0, 0) is Question 1 -0.246 0.072 -0.934 0.478 -0.914 -0.855 0.710 0.262 .

2. Answer the following questions. (A) [50%] Given the vector field F(x, y, z) = (x²y, e", yz²), verify the differential identity Vx (VF) V(V •F) - V²F (B) [50%] Remark. You are confined to use the differential identities. Let u and v be scalar fields, and F be a vector field given by F = (Vu) x (Vv) (i) Show that F is solenoidal (or incompressible). (ii) Show that G = (uvv – vVu) is a vector potential for F.

Answer 2

Question

Chapter E, Problem 1PS

To determine

Find the all values in the given table.

Expert Solution & Answer

Answer to Problem 1PS

The complete table is

Degree	Radians	Sine	Cosine	Tangent
−45° 180° −240° 210° 90°	−π4 π−4π3 7π6 π2	−12 032 −12 1	12 −1−12 −32 0	−10−3 13 undefined

Explanation of Solution

Given:

The given table is

Student's Solution and Survival Manual for Calculus, Chapter E, Problem 1PS

Calculation:

180°=πc1°=πc180−45°=πc180×−45°=−πc4sin(− 45°)=−sin( 45°)=−12cos(− 45°)=cos( 45°)=12tan(− 45°)=−tan( 45°)=−1

πc=180°sin( 180°)=0cos( 180°)=−1tan( 180°)=0

πc=180°πc= 180°π−4π3= 180°π×(− 4π3)=−240°sin(− 240°)=−sin( 240°)=−sin( 180°+ 60°)=sin( 60°)=32cos(− 240°)=cos( 240°)=cos( 180°+ 60°)=−cos( 60°)=−12tan(− 240°)=−tan( 240°)=−tan( 180°+ 60°)=−tan( 60°)=−3

180°=πc1°=πc180210°=πc180×210°=( 7π 6)csin( 210°)=sin( 180°+ 30°)=−sin( 30°)=−12cos( 210°)=cos( 180°+ 30°)=−cos( 30°)=−32tan( 210°)=tan( 180°+ 30°)=tan( 30°)=13

180°=πc1°=πc18090°=πc180×90°=( π 2)csin( 90°)=1cos( 90°)=0tan( 90°)=undefined

Hence the complete table is

Degree	Radians	Sine	Cosine	Tangent
−45° 180° −240° 210° 90°	−π4 π−4π3 7π6 π2	−12 032 −12 1	12 −1−12 −32 0	−10−3 13 undefined

Want to see more full solutions like this?

Subscribe now to access step-by-step solutions to millions of textbook problems written by subject matter experts!

Answer 3

Question

Chapter E, Problem 1PS

To determine

Find the all values in the given table.

Expert Solution & Answer

Answer to Problem 1PS

The complete table is

Degree	Radians	Sine	Cosine	Tangent
−45° 180° −240° 210° 90°	−π4 π−4π3 7π6 π2	−12 032 −12 1	12 −1−12 −32 0	−10−3 13 undefined

Explanation of Solution

Given:

The given table is

Student's Solution and Survival Manual for Calculus, Chapter E, Problem 1PS

Calculation:

180°=πc1°=πc180−45°=πc180×−45°=−πc4sin(− 45°)=−sin( 45°)=−12cos(− 45°)=cos( 45°)=12tan(− 45°)=−tan( 45°)=−1

πc=180°sin( 180°)=0cos( 180°)=−1tan( 180°)=0

πc=180°πc= 180°π−4π3= 180°π×(− 4π3)=−240°sin(− 240°)=−sin( 240°)=−sin( 180°+ 60°)=sin( 60°)=32cos(− 240°)=cos( 240°)=cos( 180°+ 60°)=−cos( 60°)=−12tan(− 240°)=−tan( 240°)=−tan( 180°+ 60°)=−tan( 60°)=−3

180°=πc1°=πc180210°=πc180×210°=( 7π 6)csin( 210°)=sin( 180°+ 30°)=−sin( 30°)=−12cos( 210°)=cos( 180°+ 30°)=−cos( 30°)=−32tan( 210°)=tan( 180°+ 30°)=tan( 30°)=13

180°=πc1°=πc18090°=πc180×90°=( π 2)csin( 90°)=1cos( 90°)=0tan( 90°)=undefined

Hence the complete table is

Degree	Radians	Sine	Cosine	Tangent
−45° 180° −240° 210° 90°	−π4 π−4π3 7π6 π2	−12 032 −12 1	12 −1−12 −32 0	−10−3 13 undefined

Want to see more full solutions like this?

Subscribe now to access step-by-step solutions to millions of textbook problems written by subject matter experts!

Answer 4

Question

Chapter E, Problem 1PS

To determine

Find the all values in the given table.

Expert Solution & Answer

Answer to Problem 1PS

The complete table is

Degree	Radians	Sine	Cosine	Tangent
−45° 180° −240° 210° 90°	−π4 π−4π3 7π6 π2	−12 032 −12 1	12 −1−12 −32 0	−10−3 13 undefined

Explanation of Solution

Given:

The given table is

Student's Solution and Survival Manual for Calculus, Chapter E, Problem 1PS

Calculation:

180°=πc1°=πc180−45°=πc180×−45°=−πc4sin(− 45°)=−sin( 45°)=−12cos(− 45°)=cos( 45°)=12tan(− 45°)=−tan( 45°)=−1

πc=180°sin( 180°)=0cos( 180°)=−1tan( 180°)=0

πc=180°πc= 180°π−4π3= 180°π×(− 4π3)=−240°sin(− 240°)=−sin( 240°)=−sin( 180°+ 60°)=sin( 60°)=32cos(− 240°)=cos( 240°)=cos( 180°+ 60°)=−cos( 60°)=−12tan(− 240°)=−tan( 240°)=−tan( 180°+ 60°)=−tan( 60°)=−3

180°=πc1°=πc180210°=πc180×210°=( 7π 6)csin( 210°)=sin( 180°+ 30°)=−sin( 30°)=−12cos( 210°)=cos( 180°+ 30°)=−cos( 30°)=−32tan( 210°)=tan( 180°+ 30°)=tan( 30°)=13

180°=πc1°=πc18090°=πc180×90°=( π 2)csin( 90°)=1cos( 90°)=0tan( 90°)=undefined

Hence the complete table is

Degree	Radians	Sine	Cosine	Tangent
−45° 180° −240° 210° 90°	−π4 π−4π3 7π6 π2	−12 032 −12 1	12 −1−12 −32 0	−10−3 13 undefined

Want to see more full solutions like this?

Subscribe now to access step-by-step solutions to millions of textbook problems written by subject matter experts!

Answer 5

Chapter E, Problem 1PS

To determine

Find the all values in the given table.

Expert Solution & Answer

Answer to Problem 1PS

The complete table is

Degree	Radians	Sine	Cosine	Tangent
−45° 180° −240° 210° 90°	−π4 π−4π3 7π6 π2	−12 032 −12 1	12 −1−12 −32 0	−10−3 13 undefined

Explanation of Solution

Given:

The given table is

Student's Solution and Survival Manual for Calculus, Chapter E, Problem 1PS

Calculation:

180°=πc1°=πc180−45°=πc180×−45°=−πc4sin(− 45°)=−sin( 45°)=−12cos(− 45°)=cos( 45°)=12tan(− 45°)=−tan( 45°)=−1

πc=180°sin( 180°)=0cos( 180°)=−1tan( 180°)=0

πc=180°πc= 180°π−4π3= 180°π×(− 4π3)=−240°sin(− 240°)=−sin( 240°)=−sin( 180°+ 60°)=sin( 60°)=32cos(− 240°)=cos( 240°)=cos( 180°+ 60°)=−cos( 60°)=−12tan(− 240°)=−tan( 240°)=−tan( 180°+ 60°)=−tan( 60°)=−3

180°=πc1°=πc180210°=πc180×210°=( 7π 6)csin( 210°)=sin( 180°+ 30°)=−sin( 30°)=−12cos( 210°)=cos( 180°+ 30°)=−cos( 30°)=−32tan( 210°)=tan( 180°+ 30°)=tan( 30°)=13

180°=πc1°=πc18090°=πc180×90°=( π 2)csin( 90°)=1cos( 90°)=0tan( 90°)=undefined

Hence the complete table is

Degree	Radians	Sine	Cosine	Tangent
−45° 180° −240° 210° 90°	−π4 π−4π3 7π6 π2	−12 032 −12 1	12 −1−12 −32 0	−10−3 13 undefined

Answer 6

Chapter E, Problem 1PS

To determine

Find the all values in the given table.

Expert Solution & Answer

Answer to Problem 1PS

The complete table is

Degree	Radians	Sine	Cosine	Tangent
−45° 180° −240° 210° 90°	−π4 π−4π3 7π6 π2	−12 032 −12 1	12 −1−12 −32 0	−10−3 13 undefined

Explanation of Solution

Given:

The given table is

Student's Solution and Survival Manual for Calculus, Chapter E, Problem 1PS

Calculation:

180°=πc1°=πc180−45°=πc180×−45°=−πc4sin(− 45°)=−sin( 45°)=−12cos(− 45°)=cos( 45°)=12tan(− 45°)=−tan( 45°)=−1

πc=180°sin( 180°)=0cos( 180°)=−1tan( 180°)=0

πc=180°πc= 180°π−4π3= 180°π×(− 4π3)=−240°sin(− 240°)=−sin( 240°)=−sin( 180°+ 60°)=sin( 60°)=32cos(− 240°)=cos( 240°)=cos( 180°+ 60°)=−cos( 60°)=−12tan(− 240°)=−tan( 240°)=−tan( 180°+ 60°)=−tan( 60°)=−3

180°=πc1°=πc180210°=πc180×210°=( 7π 6)csin( 210°)=sin( 180°+ 30°)=−sin( 30°)=−12cos( 210°)=cos( 180°+ 30°)=−cos( 30°)=−32tan( 210°)=tan( 180°+ 30°)=tan( 30°)=13

180°=πc1°=πc18090°=πc180×90°=( π 2)csin( 90°)=1cos( 90°)=0tan( 90°)=undefined

Hence the complete table is

Degree	Radians	Sine	Cosine	Tangent
−45° 180° −240° 210° 90°	−π4 π−4π3 7π6 π2	−12 032 −12 1	12 −1−12 −32 0	−10−3 13 undefined

Answer 7

Chapter E, Problem 1PS

To determine

Find the all values in the given table.

Answer 8

Expert Solution & Answer

Answer to Problem 1PS

The complete table is

Degree	Radians	Sine	Cosine	Tangent
−45° 180° −240° 210° 90°	−π4 π−4π3 7π6 π2	−12 032 −12 1	12 −1−12 −32 0	−10−3 13 undefined

Answer 9

Expert Solution & Answer

Answer 10

Expert Solution & Answer

Answer 11

Explanation of Solution

Given:

The given table is

Student's Solution and Survival Manual for Calculus, Chapter E, Problem 1PS

Calculation:

180°=πc1°=πc180−45°=πc180×−45°=−πc4sin(− 45°)=−sin( 45°)=−12cos(− 45°)=cos( 45°)=12tan(− 45°)=−tan( 45°)=−1

πc=180°sin( 180°)=0cos( 180°)=−1tan( 180°)=0

πc=180°πc= 180°π−4π3= 180°π×(− 4π3)=−240°sin(− 240°)=−sin( 240°)=−sin( 180°+ 60°)=sin( 60°)=32cos(− 240°)=cos( 240°)=cos( 180°+ 60°)=−cos( 60°)=−12tan(− 240°)=−tan( 240°)=−tan( 180°+ 60°)=−tan( 60°)=−3

180°=πc1°=πc180210°=πc180×210°=( 7π 6)csin( 210°)=sin( 180°+ 30°)=−sin( 30°)=−12cos( 210°)=cos( 180°+ 30°)=−cos( 30°)=−32tan( 210°)=tan( 180°+ 30°)=tan( 30°)=13