2 sin 2 (x) 1 = 0 2 sin 2 (x) = 1. sin 2 (x) = sin(x) = ± 2. x 3 = sin 1 ( 1/ 2) = π. x 2 = π x 1 = 3π 4. 4 x 4 = π x 3 = 5π 4

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1 Math Assignment 1 Solutions - Spring 01 - BSU - Jaimos F Skriletz 1 1. Trigonometric Equations Find all solutions to the following trigonometric equations: (a) sin (x) 1 = 0 sin (x) = 1 If sin(x) = 1/ then the two principal solutions are sin (x) = 1 1 sin(x) = ± = ± 1 = ± x 1 = sin 1 (1/ ) = π 4 x = π x 1 = 3π 4 If sin(x) = 1/ then the two principal solutions are x 3 = sin 1 ( 1/ ) = π 4 x 4 = π x 3 = 5π 4 Now since x 3 = π/4 has the same terminal point as 7π/4 we get the solutions are of the following forms x = x 1 + nπ = π 4 + nπ x = x + nπ = 3π 4 + nπ x = x 4 + nπ = 5π 4 + nπ or x = x 3 + nπ = 7π 4 + nπ Where n = 0, ±1, ±, ±3, Optional: In this case the above four solutions appear at all the 45 angles in each quadrant so we can simplify the above and write all the solutions as follows: where k = 0, ±1, ±, ±3, x = π 4 + kπ

2 Math Assignment 1 Solutions - Spring 01 - BSU - Jaimos F Skriletz (b) sin (x) cos(x) = 1 (1 cos (x)) cos(x) = 1 cos (x) cos(x) = 1 cos (x) cos(x) + 1 = 0 cos (x) + cos(x) 1 = 0 ( cos(x) 1)(cos(x) + 1) = 0 cos(x) 1 = 0 or cos(x) + 1 = 0 cos(x) = 1 or cos(x) = 1 If cos(x) = 1/ then the two principal solutions are x 1 = cos 1 (1/) = π 3 x = π x 1 = 5π 3 If cos(x) = 1 then the two principal solutions are x 3 = cos 1 ( 1) = π x 4 = π x 1 = π = x 3 Thus the general solutions are of the form x = x 1 + nπ = π 3 + nπ x = x + nπ = 5π 3 + nπ or x = x 3 + nπ = π + nπ = (n + 1)π Where n = 0, ±1, ±, ±3,

3 Math Assignment 1 Solutions - Spring 01 - BSU - Jaimos F Skriletz 3 (c) sin(x) cos(x) = 4 sin(x) cos(x) = sin(x) = Let u = x then we want the solutions to sin(u) = /. The two principal solutions are u 1 = sin 1 ( /) = π 4 u = π u 1 = 3π 4 Thus the general solutions are u = x = u 1 + nπ = π 4 + nπ or u = x = u + nπ = 3π 4 = nπ Divide everything by to find x to get x = π 8 + nπ or x = 3π 8 + nπ Where n = 0, ±1, ±, ±3,

4 Math Assignment 1 Solutions - Spring 01 - BSU - Jaimos F Skriletz 4 (d) cos(x) sin(x) = 1 First we can write sin(x) + cos(x) as a single sine function by finding the polar coordinates of (A, B) = ( 1, 1). First the distance is r = A + B =. Second (A, B) = ( 1, 1) is located in quadrant II with a reference angle of φ = π/4, thus φ = 3π/4. Put this together to get that sin(x) + cos(x) = r sin(x + φ) = sin(x + 3π/4) Last let u = x + 3π/4 and simplify the equation as follows The two principal solutions are sin(x) + cos(x) = 1 sin(x + 3π/4) = 1 sin(u) = 1 u 1 = sin 1 (1/ ) = π 4 The general solutions are Solve the above for x to get the solutions are u = π u 1 = 3π 4 u = x + 3π 4 = u 1 + nπ = π 4 + nπ or u = x + 3π 4 = u + nπ = 3π 4 + nπ x = π + nπ = π 4 + nπ or x = 0 + nπ = nπ

5 Math Assignment 1 Solutions - Spring 01 - BSU - Jaimos F Skriletz 5. Polar Coordinates (a) Convert the following polar coordinates to Cartesian coordinates then plot them on the plane showing both the Cartesian and polar coordinates in the plot. i. (r, θ) = (7, 4π/3) The rectangular coordinates for this point are x = r cos(θ) = 7 cos(4π/3) = 7( 1/) = 7/ y = r sin(θ) = 7 sin(4π/3) = 7( 3/) = 7 3/ So the rectangular coordinates are (x, y) = ( 7/, 7 3/) as shown: ii. (r, θ ) = ( 10, 3π/6) The rectangular coordinates for this point are x = r cos(θ) = ( 10) cos(3π/6) = ( 10)( 3/) = 5 3 y = r sin(θ) = ( 10) sin(3π/6) = ( 10)( 1/) = 5 So the rectangular coordinates are (x, y) = ( 5 3, 5) as shown:

6 Math Assignment 1 Solutions - Spring 01 - BSU - Jaimos F Skriletz 6 (b) Convert each of the following Cartesian Coordinates to Polar Coordinates: i. (x, y) = ( 6, 6) Plot the point on the plane as shown: The distance the point is from the origin is r = x + y = ( 6) + ( 6) = 7 r = ± 7 = ±6 The reference angle is found by tan(θ) = y/x = ( 6)/( 6) = 1 The reference angle is θ = π/4 and is located in Quadrant III, thus θ = 5π/4 Putting this together we get the following list of polar coordinates, (r, θ), for r = 6 > 0: = (6, 11π/4) = (6, 3π/4) = (6, 5π/4) = (6, 13π/4) = (6, 1π/4) = And we get the following for r = 6 < 0 = ( 6, 15π/4) = ( 6, 7π/4) = ( 6, π/4) = ( 6, 9π/4) = ( 6, 17π/4) = ii. (x, y) = ( 1, 0) Plot the point on the plane as shown: Points on the axis are fairly straight forward, first notice that the distance from the origin is r = 1 = 1 Thus we get the following list of polar coordinates, (r, θ), for r = 1 > 0 And the following for r = 1 < 0 And that the angle of rotation to the negative x-axis is θ = π = (1, 3π) = (1, π) = (1, π) = (1, 3π) = (1, 5π) = = ( 1, 4π) = ( 1, π) = ( 1, 0) = ( 1, π) = ( 1, 4π) =

7 Math Assignment 1 Solutions - Spring 01 - BSU - Jaimos F Skriletz 7 3. Graphs of Polar Equations Plot the graph to the following polar equations: (a) r = 5 sin(3θ) (b) r = 1 3 cos(θ)

8 Math Assignment 1 Solutions - Spring 01 - BSU - Jaimos F Skriletz 8 4. Polar form of Complex Numbers Use the polar form of complex numbers to calculate ( ) i First convert the point (x, y) = (1/, 3/) into polar coordinates. To do this notice the distance from the origin is Thus the distance is r = 1 from the origin. r = (1/) + ( 3/) = 1/4 + 3/4 = 1 Next notice that the reference angle is theta = π/3 and is quadrant IV so θ = 5π/3. Thus we have that Thus the 15th power is Last convert back into rectangular form to get z = i = 1[cos(5π/3) + i sin(5π/3)] = cis(5π/3) z 15 = [cis(5π/3)] 15 = cis(15(5π/3)) = cis(5π) = cos(5π) + i sin(5π) ( ) i = cos(5π) + i sin(5π) = 1 + i(0) = 1

9 Math Assignment 1 Solutions - Spring 01 - BSU - Jaimos F Skriletz 9 5. Roots of Complex Numbers Use polar coordinates for complex numbers to answer the following: (a) Find all three cube roots of z = i. First write the point (x, y) = ( 4 3, 4) in polar form: The distance this point is from the origin is r = ( 4 3) + (4) = 16(3) + 16 = 64 Thus r = 64 = 8. Next the angle is in quadrant II and solves the equation Thus the angle is of the form θ = 5π/6 So we can write this complex point as The cube roots of this angle are: tan(θ) = y x = = 1 3 z = i = 8cis(5π/6) = 8cis(5π/6 + nπ) z 1/3 = [8cis(5π/6 + nπ)] 1/3 = 8 1/3 cis((1/3)[5π/6 + nπ]) = cis(5π/18 + nπ/3) So the principal root is z 1 = cis(5π/18) = cos(5π/18) + i sin(5π/18) And the other roots can be found by adding a third of a rotation π/3 to the previous root as shown z 1 = cis(5π/18) = cos(5π/18) + i sin(5π/18) z = cis(5π/18 + π/3) = cis(17π/18) = cos(17π/18) + i sin(17π/18) z 3 = cis(5π/18 + (π/3)) = cis(9π/18) = cos(9π/18) + i sin(9π/18)

10 Math Assignment 1 Solutions - Spring 01 - BSU - Jaimos F Skriletz 10 (b) Find all four forth roots of z = i. First write the point (x, y) = (0, 1) in polar form: The polar form of this point will have a distance r = 1 from the origin and have the angle θ = 3π/ (negative y-axis): Thus we can write the point as So the forth roots are then z = i = 1[cos(3π/) + i sin(3π/)] = cis(3π/) = cis(3π/ + nπ) z 1/4 = [cis(3π/ + nπ)] 1/4 = cis((1/4)[3π/ + nπ]) = cis(3π/8 + nπ/) So the principal root is z 1 = cis(3π/8) = cos(3π/8) + i sin(3π/8) And the other roots are found by adding a quarter of rotation, π/, to the previous roots. This gives us all four roots as shown: z 1 = cis(3π/8) = cos(3π/8) + i sin(3π/8) z = cis(3π/8 + π/) = cis(7π/8) = cos(7π/8) + i sin(7π/8) z 3 = cis(3π/8 + π/) = cis(11π/8) = cos(11π/8) + i sin(11π/8) z 4 = cis(3π/8 + 3π/) = cis(15π/8) = cos(15π/8) + i sin(15π/8)

Chapter 5 Notes. 5.1 Using Fundamental Identities

Chapter 5 Notes. 5.1 Using Fundamental Identities Chapter 5 Notes 5.1 Using Fundamental Identities 1. Simplify each expression to its lowest terms. Write the answer to part as the product of factors. (a) sin x csc x cot x ( 1+ sinσ + cosσ ) (c) 1 tanx