MATH 4 part 2

Lesson 4 Find the values of six trigonometric functions for /A given some conditions You can determine five other trigonometric functions if one function isgiven.Example 1. if sin A = 4/5, A is not in QI find the other function values for A. Solution: Since sin function is positive in QI & II and /A is not in QI as stated in the given , so /A is in QII. Consider the algebraic sign of five other functions in QII. Since Sin A = y/r and y = 4; r = 5 Solve for x Figure B 45 9Find x using Pythagorean theorem r2 = x2 + y2 r2 - y2 = x2 52 -42 = x2 25 –16 = x 9 = x2 +3 = xsince A is in QII the value of x =-3Now the ratios are:Cos A = -3/5 Csc A = 5/4Tan A = -4/3 Sec A = -5/3Cot A = -3/4Example 2. Tan A = √3/3, cos A < 0 and sin A <0 , find values of five other trigonometric function for / A.

Solution: Terminal point lies in quadrant III. Let us consider the algebraic sign of the function in QIII. Tan and Cot are the only positive function while the rest are negative . Tan A = y/x y = -√3 ; x = -3 Figure X = -3 -√3 = ySolve for r:By Pythagorean theoremr2 = x2 + y2r2 = - 32 +(-√3)2r2 = 9 + 3r2 = 12r = 2√3 The function ratios are: 4. Sec A = -2√3/31. Sin A = - ½ 5. Cot A = √32. Cos A = - √3/23.Csc A = -2Example 3: If sin A = √7/4 , and π/2 < A < π. Find the values of five other trigonometric functions for A.Solution: Sin A = √7/4 and we define sin A = y/r y = √7; r = 4 and A lies in QII , sin A and csc A are positive while five other function are negative. . Figure r=4 y =√7 xYou need to find xr2 = x2 + y2

r2 –y2 = x242 - √72 = x242- √72 = x216 – 7 = x29 = x2-3 = xTrigonometric ratios are:cos A = x /r = -3/ 4Tan A = y/x =- √7/3Csc A = r/y = 4 = 4√7 rationalize √7 (√7)(√7) = 4√7 7Sec A = r/x = 4/-3Cot A = x/ y = -3 √7Example 4: If Sec B = 8/3 and 0 < B < 90, find the other function values for B.Solution: Since sec B and cos B are reciprocals cos B = 3/8 We define cos B = x/r, so x = 3 ; r = 8; B lies in QI, all the functions have positive sign.Figure 8 X =3You need to solve for y r2 = x2 + y2 r2 – x2 = y2 82 – 32 = y2 64 - 9 = y2 55 = y2

√55 = yTrigonometric ratios are:Sin B = y/r = √55/8Tan B = y/x = √55/3Csc B = r/y = 8 = 8√55 rationalize √55 (√55)(√55) = 8√55 55Cot B = x/y = 3 = 3√55 rationalize √55 (√55)(√55) = 3√55 55Try this outA. Find the value of each of the remaining functions of the acute angle A :1.If Cos A = 5/13 and Sin A < 0 Figure Find: y= d. Csc A =a. Sin A =b. Tan A = e. Cot A =c. Sec A =2. IF Sin A = √3/4 and π/2 < A <π Figure Find: x=a Cos A = d. Sec A =b. Tan A = e. Cot A =c. Csc A =3. If Sin A = 2/3 and cos A > 0 Figure Find: . x=

a. Cos A = d. Csc A =b. Tan A = e. CoT A =c. Sec A =4. If Cos A = -2√2/3 and Tan A > 0 Figure Find: Figure y=a. Sin A = d. Sec A =b. Tan A = e. Cot A =c. Csc A =5. If Tan A = -3/4 ,270 < A< 360 Find: r=a. Sin A = d. Csc A =b. Cos A = e. Sec A =c. Cot A =What have you learnedGiven the following function, identify the amplitude of: 1. y = ½ cos x 2. y = 4 cos x 3. y = sin 4xDetermine the period of the following functions. 4. y = 3 sin x 5. y = cos x 6. y = 4 sin 4x 7. What is the value of y = ½ cos x if x = π/2

Given the figure, find the value of r and the six trigonometric functions. -7 A -24 r 8.r = 9. sin A = 10. Cos A = 11. Tan A = 12. Cot A = 13 .Csc A = 14.sec A =Sketch the figure, find the values of r and 5 other trigonometric functions for /C ifTan C = -4/3 and Sin C > 0. 15. r = 16. sin C = 17. Cos C = 18. Cot C= 19. Sec C = 20. Csc C =

Answer key Try this outHow much do you know1. y =22. y = ¾3. y = -24. 4π5. 3π6. 10π7. sol: y = 4 sin ½(5π/3) y = 4 sin 5π/6 y = 4(½) y =28. r = 139. sin A = 12/1310. Cos A = -5/1311. Cot A = -5/1212. Csc A = 13/1213. Tan A = -12/514. Sec A = -13/515. r = 1016. Sin k = -3/517. Cos S = 4/518. Cot S= -4/319. Csc S = -13/1220. Sec S = 5/4Lesson 1 A. 1. amplitude : 3 2. amplitude 2 3. amplitude: 3/2 4. amplitude: 2 5 amplitude: 1 6. amplitude: 2 7. amplitude: ½ 8. amplitude: ¾ 9. amplitude: 4 10. amplitude: 2 B. 1. Period: 3π 2. Period: 2π 3. Period 4π

4. Period: 8π/3 5. Period : 4π 6. Period : 2π/3 7 Period : 3π 8. Period: π 9. Period: 10π 10. Period: π/2Lesson 2. Table of values 1.x 0 π/4 π/2 3π/4 π 5π/4 3π/2 7π/4 2πY =1/2 cos x .5 .4 0 -.4 -.5 -.4 0 .4 .5 Solution: b. If x = π/4 c. If x = π/2 d. If x = 3π/4 a. If x = 0 y = ½ cos π/4 y = ½ cos π/2 y = ½ cos 3π/4 y = ½ cos 0 y = ½ (√2/2) y = ½ (0) y = ½ (-√2/2) y = ½ (1) y = .4 y=0 y = -.4 y = .5 e. If x = π f. If x = 5π/4 g. If x = 3π/2 h. If x = 7π/4 y = ½ cos π y = ½ cos 5π/4 y = ½ cos 3π/2 y = ½ cos 7π/4 y = ½ (-1) y = ½ (-√2/2) y = ½ ( 0) y = ½ (√2/2) y = -.5 y = -.4 y=0 y = .4 i. If x = 2π y = ½ cos 2π y = ½ (1) y = .5 Graph of y = ½ cos x π/4 π/2 3π/4 π 5π/4 3π/2 7π/4 2π 4. y = 4 sin ½ x, where 0 ≤ x ≤ 2π

Table of Values:x 0 π/3 π/2 2π/3 π 4π/3 3π/2 5π/3 2πy = 4 sin 1/2x 0 2 2.83 3.46 4 3.46 2.83 2 0 Solution: b. If x = π/3 c. If x = π/2a. If x = 0 y = 4 sin ½ (π/3) y = 4 sin ½ (π/2) y = 4 sin (π/6) y = 4 sin (π/4) y = 4 sin ½(0) y = 4 (1/2) y = 4 (√2/2) y = 4 sin 0 y=2 y = 2.83 y = 4 (0) y=0 a. If x = 2π/3 e. If x = π f. If x = 4π/3 y = 4 sin ½(2π/3) y = 4sin ½(π) y = 4 sin ½(4π/3) y = 4 sin π/3 y = 4(1) y = 4 sin 4π/6 y = 4(√3/2) y=4 y = 4 sin 2π/3 y = 2√3 = 3.46 y = 3.46 h. If x = 5π/3g. If x = 3π/2 y = 4 sin ½ (5π/3) i. If x = 2π y = 4 sin ½ (3π/2) y = 4(5π/6) y = 4 sin ½(2π) y = 4 sin 3π/4 y = 4(½) y = 4 sin π y = 4 (-√2/2) y=2 y = 4(0) y = 2.83 y=0 Graph of y = 4 sin ½ x π/3 π/2 2π/3 π 4π/3 3π/2 5π/3 2π3. y = 2/3 cos x, where 0 ≤ x ≤ 2π Table of values:

x 0 π/4 π/2 3π/4 π 5π/4 3π/2 7π/4 2π y = 2/3 cos x .7 .47 0 -.47 -.7 -.47 0 .47 .7Solution: b. If x = π/4 c. If x = π/2 d. If x = 3π/4a. If x = 0 y = 2/3 cos π/4 y = 2/3 cos π/2 y = 2/3 cos 3π/4y = 2/3 cos 0 y = 2/3 (√2/2) y = 2/3 (0) y = 2/3 (-√2/2)y = 2/3 (1) y = .47 y=0 y = -.47y = .7f.) If x = π g. If x = 5π/4 h. If x = 3π/2 i. If x = 7π/4y = 2/3 cos π y = 2/3 cos 5π/4 y = 2/3 cos 3π/2 y = 2/3 cos 7π/4y = 2/3 (-1) y = 2/3 (-√2/2) y = 2/3 (0) y = 2/3 (√2/2)y = -.7 y = -.47 y=0 y = .47j. If x = 2π y = 2/3 cos 2π y = 2/3 (1) y = .7 Graphs of y = 2/3 cos x.5.5 π/4 π/2 3π/4 π 5π/4 3π/2 7π/2 2π4.y = 2 cos 1/2x where 0 < x < 2π Table of values: x0 π/3 π/2 2π/3 π 4π/3 3π/2 5π/3 2πy = 2 cos1/2 x 2 1 -√2 -√3 -2 √3 √2 1 0

Solution:a. x = 0 b. x = π/3 c. x = π/2 d. x = 2π/3y = 2 cos ½(0) y = 2 cos ½(π/3) y = 2 cos ½(π/2) y = 2 cos ½(2π/3)y = 2 Cos 0 y = 2 cos π/6 y = 2 cos π/4 y = 2 cos 2π/6y = 2(1) y = 2(√3/2) y = 2(√2/2) y = 2 cos π/3y=2 y = √3 y = √2 y = 2(½) y=1e. x = π f. x = 4π/3 g. x = 3π/2 h. x = 5π/3y = 2 cos ½(π) y = 2 cos ½(4π/3) y = 2 cos ½(3π/2) y = 2 cos ½(5π/3)y = 2 cos π/2 y = 2 cos 4π/6 y = 2 cos 3π/4 y = 2 cos 5π/6y = 2(0) y = 2 cos 2π/3 y = 2(√2/2) y = 2(√3/2)y=0 y = 2(½) y = -√2 y = -√3 y=1i. x = 2πy = 2 cos ½(2π)y = 2 cos πy = 2(-1)y = -25. y = ¾ sin ½ x Table of values x0 π/3 π/2 2π/3 π 4π/3 3π/2 5π/3 2πy = ¾ sin ½ x 0 3/8 3√2/8 3√3/8 ¾ 3√3/8 3√2/8 3/8 0Solution: b. x = π/3 c. x = π/2 d. x = 2π/3a. x = 0 y = ¾ sin ½(π/3) y = ¾ sin ½(π/2) y = ¾ sin ½(2π/3) y = ¾ sin ½(0) y = ¾ sin 0 y = ¾ sin π/6 y = ¾ sin π/4 y = ¾ sin π/3 y = ¾(0) y=0 y = ¾(½) y = ¾(√2/2) y = ¾(√3/2) y = 3/8 y = 3√2/8 y = 3√3/8e. x = π f. x = 4π/3 g. x = 3π/2 h. x = 5π/3 y = ¾ sin ½(π) y = ¾ sin π/2 y = ¾ sin ½(4π/3) y = ¾ sin ½(3π/2) y = ¾ sin ½(5π/3) y = ¾(1) y=¾ y = ¾ sin 2π/3 y = ¾ sin 3π/4 y = ¾ sin 5π/6 y = ¾(√3/2) y = ¾(√2/2) y = ¾(½) y = 3√3/8 y = 3√2/8 y = 3/8i. x = 2π y = ¾ sin ½(2π) y = ¾ sin π y = ¾(0) y=0

Lesson 3Try this out: 1. (-5, 7)Figure 7r -5 c. Tan A = -7/5 e. Cot A = -5/7Solve for rBy Pythagorean Theorem d. Sec A = √74 f. Csc A = √74 5 7 r2 = (x) 2 + (y)2 = (5) 2 + (7) 2 Solve for r = 25 + 49 By Pythagorean Theorem: r = √74 a. Sin A = 7√74 r2 = (x) 2 + (y) 2 74 = (-8) 2 + (-15) 2 b. Cos A = 5√74 = 64 + 225 742. (-8, -15) = 289 -8 r = 17-15 Sin A = -15/17 Csc A = -17/15 2. (24, - 7) Cos A = -8/17 Sec A = -17/8 figure Tan A = 15/8 Cot A = 8/15 By Pythagorean Theorem: 24 r2 = x2 + y2 = (24) 2 + (-7) 2 -7 = 576 + 49 = 625

r = 25 Sin B = -7/25 Csc B = -25/7 Cos B = 24/25 Sec B = 25/24 Tan B = -7/24 Cot B = -24/73. (2, 3) Figure Solve for r By Pythagorean theorem: 3 r2 = x2 +y2 2 = 22 + 32 =4+9 = 13 r = √13 Sin A = 3√13/13 Csc A = √13/ Cos A = 2√13/13 Sec A = √13/2 Tan A = 3/2 Cot A = 2/34. (-9, 40) Figure40 Solve for r By Pythagorean Theorem -9 r2 = x2 + y2 = (-9) 2 + (40) 2 = 81 + 1600 = 1681 r = 41 Sin A = 40/41 Csc A = 41/40 Cos A = -9/41 Sec A = -41/9 Tan A = 40/-9 Cot A = -9/40Lesson 3 1. Cos A = 5/13, If A is in Q IV Sol: In Q IV the only positive functions are cos A and sec A .Sketch: By Pythagorean Theorem x2 + y2 = r2 5 y2 = (13) 2 – (5) 2 13 y2 = 169 – 25 y2= 144 y = 12

a. Sin = -12/13 d. Csc A = -13/12b. Tan A = -12/5 e. Cot A = - 5/12c. Sec A = 13/52. Sin A = √3/4, If A is in Q II Sol: In Q II sin A and CSC A are positive and the rest of the functions are negative. By Pythagorean Theorem x2 + y2 = r2 √3 4 x2 = r2 – y2 x2 = 42 – (√3) 2 d. Sec A = -4√13/13 x2 = 16 – 3 e. Cot A = -√39/3 x = + √ 13 , since A is in QII x = -√13 a. Cos A = - √13/4 b. Tan A = - √39/13 c. CsC A = - 4√3/33. Sin A = 2/3, Where A is in Q ISol: Since A is in Q I, all the functions are positive. 32 By Pythagorean Theorem d. Csc A = 3/2 x2 + y2 = r2 e. CoT A = √5/2 x2 = (3)2 – (2)2 x2 = 9 – 4 x2 = 5 x = +√5 but A is in Q I then x =+√5 a. Cos A = √5/3 b. Tan A = 2√5/5 c. Sec A = 3√5/54. Cos A = -2√2/3, A is in QIII

Sol: since a is in QIII, tan and cot are the only positive and the rest are negative. -2√2 3By Pythagorean theorem d. Sec A = 2√2/3x2 + y2 = r2 e. Cot A = 2√2y2 = (3) 2 – (2√2) 2 =9–8 y=1 a. Sin A = -1/3 a. Tan A = √2/4 b. Csc A = -35. Tan A = -3/4, A is in QIV Sol: Since A is in QIV, cos and sec are the only positive and the rest are negative. 3 By Pythagorean Theorem -4 r2 = x2 + y2 r2 = (4) 2 + (-3)2 = 16 + 9 = 25 r=5a. Sin A = -3/5 d. Csc A = -5/3b. Cos A = 4/5 e. Sec A = -5/4c. Cot A = -4/3Post Test1. y = 1/22. y = 13. 44. 2π5. π4.6. π/2

7. y = 0 8. r = 259. Sin A = -24/2510. Cos A = -7/2511. Tan A = 24/712. Csc A = -25/2413. Sec A = -25/714. Cot A = 7/2415. r = 516. sin C = -4/517. Tan C = -4/318. Cot C = -3/419. Sec C = 5/320. Csc C = -5/4

Module 4 Quadratic Functions What this module is about This module is about the application of quadratic equations and functionsin everyday situations. As you go over the different problems you will apply yourknowledge and skills related to quadratic equations and functions in solvingproblems. Frequently, only one solution of the equation is relevant to the problemat hand. The root which does not satisfy the conditions of the problem must berejected. It is always a good practice, then, to check the solutions to determine ifone or both may be used. What you are expected to learnThis module is designed for you to:1. recall the different steps in solving word problems2. translate verbal statements into symbols3. apply knowledge and skills related to quadratic equations and functions in solving problems.How much do you knowA. Write the letter of the correct answer.1. The roots of a quadratic equation are 8 and -8. What is the equation?a. x2 + 64 = 0 b. x2 – 64 = 0 c. x2 – 8 = 0 d. x2 + 8 = 02. What are the values of a, b, and c in the equation 6x = 2x2 + 1a. a = -2 b = -6 c = 1

b. a = 2 b = -6 c. = -1c. a = -2 b=6 c. = 1d. a = 2 b = -63. Which of the following is a solution of ( x – 3 ) ( x – 4 ) = 20a = -5 b=4 c=8 d = 204. Which of the following has roots of -3 and 7?a. x2 + 4x – 21 = 0 b. x2 – 4x – 21 = 0c. x2 – 4x + 21 = 0 d. x2 + 4x + 21 = 05. Find the product of the roots of 2x2 = x + 3a. 1 b. 3 c. -1 d. − 3 22 2B. Write an equation to show the functional relationship between the quantities involved in the problems using the indicated variables.1. The area (A) of a square of side s is s2.2. The perimeter of a rectangle is equal to twice the Area.3. 12x plus seven is equal to 3x reduced by 44. The circumference (C) of a circle is twice the product of π and the radius (r).5. The total distance (d) covered is equal to the product of the rate (r) and the time (t).C. Solve the following problems.1. The product of two consecutive odd integers is 143. Find the integers.2. One side of a rectangle is 10 cm longer than its width. The area of the rectangle is 96 sq. cm. Find the dimensions of the rectangle.3. The sum of two numbers is 22 and the sum of their squares is 250. Find the numbers. 2

4. The sum of two numbers is 24. What is the maximum possible product of the two numbers? What you will do Lesson 1 Number Problems Many students could not solve this type of problem easily because theyfail to translate correctly the different expressions given by the word problem intoa correct equation. In the following examples, you will see that before you solve aword problem, you must first translate the word problem into an equation.Steps in Solving word problems involving quadratic equation and function: 1. Identify your unknowns. 2. Write your equation. 3. Operate on the numbers. 4. Convert the equation into the standard form. Remember to change signs. 5. Factor the equation into its two corresponding linear equations. By factoring, you reduce the quadratic equation into two equivalent linear equations.Examples: 1. Find two consecutive odd integers whose product is 195. Solution: 1st step: Identify your unknowns. Let x = one odd number, so x + 2 = next consecutive odd number. 3

2nd step: Write your equation. x ( x + 2 ) = 195 one odd consecutive their number odd number product3rd step: Operate on the numbers. Multiply the two quantities. Thus: x2 + 2x = 1954th step: Bring all terms to the left side of the equal sign and equate to zero. Remember to change signs.So x2 + 2x - 195 = 05th step: Factor the equation.Thus: (x + 15) (x – 13) = 06th step: Solve for x. x + 15 = 0 or (x – 13) = 0 x = -15 or x = 13The consecutive numbers are : x = 13 and x + 2 = 15, 13 and 15or x = -15 and x +2 = -15 + 2 = -13 , -13 and -15Check: or x ( x + 2) = 195 x ( x + 2) = 195 (13)(15) = 195 (-13)(-15) = 195 195 = 195 195 = 195 4

2. The sum of two numbers is 19 and their product is 60. Find the numbers.Solution:1st step: Identify the unknown. Let x = one number So 19 - x = the other numberAnd x(19 – x) = their product.2nd step: Write the equation.x ( 19 – x ) = 603rd step: Operate the numbers to remove the parentheses grouping symbol.19x – x2 = 604th step: Convert the equation to its standard form.Thus: or x2 - 19x + 60 = 0 19x - x2 - 60 = 05th step: Factor the equation. ( x – 15 ) ( x - 4 ) = 06th step: Solve for x.x – 15 = 0 and x – 4 = 0 x = 15 x =419 – x 19 – 15 = 4 19 – x 19 –4 = 15The numbers are therefore 15 and 4Check: 15 + 4 = 19 Their sum: 19 = 19 check.Their product: (15) (4) = 60 60 = 60 check. 5

3. The sum of two numbers is 22 and the sum of their squares is 250. Find the numbers.Solution:1st step: Identify your unknown Let x = one of the two numbers 22 – x = the other number2nd step: Write the equationx2 + ( 22 – x )2 = 2503rd step: Operate on the numbers to remove the parentheses grouping symbol.x2 + ( 22 – x )2 = 250x2 + 484 - 44x + x2 = 2502x2 - 44x + 484 = 2504th step: Convert the equation to its standard form.2x2 - 44x + 484 - 250 = 0 2x2 - 44x + 234 = 0 divide the equation by 2 x2 - 22x + 117 = 05th step: Factor the quadratic equation to obtain the two equivalents linear equations.( x - 13 ) ( x - 9 ) = 06th step: Solve for x.x - 13 = 0 and x - 9 = 0 x = 13 and x=9If x = 13, 22 – 13 = 9. 6

If x = 9, 22 - 9 = 13. The numbers are 9 and 13. Check: 9 + 13 = 22; 92 + 132 = 81 + 169 = 250.Try this outAnalyze and solve. 1. Find 2 consecutive integers such that three times the square of the first is equal to seven more than five times the second. 2. Find 2 consecutive even integers whose product is 224. 3. Find an integer such that the square of the integer is eighty one less than eighteen times the integer. 4. One number is three times another and their product is equal to their sum increased by fifty five. Find the numbers. 5. The sum of two numbers is 16 and the sum of their squares is 146. Find the numbers. 6. A number is 16 less than its multiplicative inverse. Find the number. 15 7. Find two numbers whose difference is 5 and the difference of their squares is 65. 8. One number is three more than a second number. The sum of their squares is 37 more than the product of the numbers. Find the two numbers. 9. The product of two consecutive integers is 47 more than the next consecutive integer. Find the two numbers. 10. The tens digit of a certain number is 7 less than the units digit. The sum of the squares of the two digits is 85. Find the numbers. 7

Lesson 2 Geometry Problems Many government and classroom examinations deals with problemsinvolving measurements of different geometric figures. These problems are alsocalled geometry problems.Examples: 1. The altitude of a triangle is 3 cm. less than the base. The area of the triangle is 35 square centimeters. What are its dimensions? Solution: 1st step: Identify the unknown. Let x cm = length of the base So (x – 3) cm = length of the altitude Area of a triangle = (base)(altitude) 2 2nd step: Write the equation. 35 = x(x − 3) 2 3rd step: Operate the numbers to remove the parentheses grouping symbol. x ( x – 3 ) = 70 x2 - 3x = 70 4th step: Convert the equation to its standard form. x2 - 3x - 70 = 0 5th step: Factor the equation. (x – 10)(x + 7) = 0 8

6th step: Solve for x.x – 10 = 0 and x + 7 = 0 x = 10 x = -7 We can not accept -7 as a value for the base of the triangle. Therefore thebase of the triangle is 10 cm while the altitude is x - 7 = 7 cm.2. The perimeter of a rectangle is 138 meters and the area is 1080 square meters. Find the length and the width of the rectangle.Solution:1st step: Identify the unknown. Let x cm = length of the rectangle 138 cm = Perimeter P = 2l + 2w 138 = 2x + 2w express perimeter in terms of x. 138 = 2(x + w) 69 = x + w w = 69 – x the widthArea of a rectangle = length x width2nd step: Write the equation. 1080 = x ( 69 – x )3rd step: Operate the numbers to remove the parentheses grouping symbol. x ( 69 - x ) = 1080 69x - x2 = 10804th step: Convert the equation to its standard form.x2 - 69x + 1080 = 05th step: Factor the equation.( x – 45) ( x - 24 ) = 0 9

6th step: Solve for x.x – 45 = 0 and x - 24 = 0 x = 45 x = 24 x = 45 m is the length of the rectangle69 – 45 = 24 m is the widthPerimeter = 2( 45 ) + 2(24) = 138 m Area = 45( 24 ) = 1080 square meters3. If the hypotenuse of a right triangle measures 13 cm and one leg is 7 cm more than the other, what are the lengths of the two legs?Solution:1st step: Identify the unknown. Let x cm = length of shorter leg (x + 7 ) cm = length of other legFormula: a2 + b2 = c22nd step: Write the equation. x2 + ( x + 7 )2 = 1333rd step: Operate the numbers to remove the parentheses grouping symbol.x2 + x2 + 14x + 49 = 169 2x2 + 14x + 49 = 1694th step: Convert the equation to its standard form.2x2 + 14x + 49 - 169 = 0 2x2 + 14x - 120 = 0simplify by dividing by 2 x2 + 7x - 60 = 05th step: Factor the equation. ( x + 12 ) ( x - 5 ) = 0 10

6th step: Solve for x.x + 12 = 0 and x - 5 = 0 x = -12 x =5 Disregard -12 as a solution because it has no real meaning. You cannothave a length of negative 12. x = 5 m is the length of of shorter legx + 7 = 12 m is the length of the other legCheck: a2 + b2 = c2 52 + (5 + 7)2 = 132 25 + 144 = 169 169 = 169Try this outSolve: 1. The length of a rectangular floor is twice the width. The area of the floor is 32 m2. What are the dimensions of the room? 2. If the perimeter of a rectangular garden is 76m and the area is 360 square meters, what are the dimensions of the garden? 3. The base of a triangle is 7 cm more than its height. If the area is 294 square centimeters, find the base and the height of the triangle. 4. A square pool was surrounded by 3 meters wide Bermuda grass. If the total area of the sidewalk and the square pool is 196 sq. m , how long is each side of the pool? What is the area of the Bermuda grass sidewalk? 5. A rectangular flower garden with dimensions 3m by 7m is surrounded by the walk of uniform width. If the area of the walk is 11 sq. m, what is the width in meters? 11

Lesson 3 Motion Problems Motion problems deal with three quantities. They are: Distance Rate or speed Time All uniform motion problems are tied-up with the formula: Distance = Time x Rate or D = rtExamples: 1. A car travels 10 kilometers per hour faster than a truck. The car goes 600 kilometers in 5 hours less time than it takes the truck to travel the same distance. Find the rate of each vehicle in kilometers per hour. Solution: 1st step: Identify the unknown.Truck Rate Time Distance Car x 600 600 600 x + 10 x 600 x + 102nd step: Write the equation.600 = 600 + 5 x x + 103rd step: Operate the numbers to remove the parentheses grouping symbol. Multiply both sides by the least common denominator which is x( x + 10 )600( x + 10 ) = 600x + 5( x2 + 10x ) 12

600x + 6000 = 600x + 5x2 + 50x4th step: Convert the equation to its standard form.Thus:5x2 + 600x - 600x + 50x - 6000 = 0Simplify by dividing the equation by 5 x2 + 10x - 1200 = 05th step: Factor the equation. (x + 40) (x - 30) = 06th step: Solve for x.x + 40 = 0 and x – 30 = 0 x = -40 x = 30 We must disregard the negative value. Thus, the rate of the truck is 30kilometers per hour. The rate of the car is x + 10 = 40 kilometers per hour.2. A man drives 500 km to a business convention. On the return trip, he increases his speed by 25 km per hour and saves 1 hour of driving time. How fast did he go in each direction?Solution:1st step: Identify the unknown. Let r = speed in going to the convention r + 25 = speed of the return trip 500 = length of time to the convention r 500 = length of time in returning from the convention r + 25 13

(length of time in going) = (length of time in returning) + 12nd step: Write the equation. 500 = 500 + 1 r r + 253rd step: Operate the numbers to remove the parentheses grouping symbol. Multiply both sides by the least common denominator which is r( r + 25 ) 500( r+ 25 ) = 500r + r( r + 25 )500r + 12, 500 = 500r + r2 + 25r500r + 12,500 = 525r + r24th step: Convert the equation to its standard form. Thus: 0 = r2 + 525r - 500r - 12,5005th step: Factor the equation.0 = ( r + 125 ) (r - 100 )6th step: Solve for x.r + 125 = 0 or r – 100 = 0r = -125 r = 100 We must disregard the negative value. Thus, the rate of the man is 100kilometers per hour in going to the convention and returns at 100 + 25 = 125kilometers per hour.2. A man can row 8 kilometers downstream and back in 6 hours. If the rate of the stream is 1 kilometer per hour, what is the rate of the man rowing in still water? 14

Solution:1st step: Identify the unknown. Let x = rate of the man in still waterDownstream D r t Upstream 8 x+1 8 x-1 8 x +1 8 x −12nd step: Write the equation. 8 + 8 =6 x +1 x −13rd step: Operate the numbers to remove the parentheses grouping symbol. Multiply both sides by the least common denominator which is ( x + 1) (x - 1)8(x – 1) + 8(x + 1) = 6(x + 1(x – 1)8x – 8 + 8x + 8 = 6(x2 – 1) 16x = 6x2 – 64th step: Convert the equation to its standard form.Thus: 0 = 6x2 - 16x - 6Simplify by dividing the equation by 2 0 = 3x2 - 8x - 35th step: Factor the equation. 0 = (3x + 1)( x - 3) 15

6th step: Solve for x. or x– 3 = 0 3x + 1 = 0 x =3 x = −1 3 We must disregard the negative value. Thus, the rate of the man is 3kilometers per hour.Try this outAnswer the following problems: 1. Two cyclist A and B traveled the same distance of 120 kilometers. A traveled 4 kilometers per hour faster than B and covered the distance in one hour less than B. Find the rate of each. 2. If Mr. Cruz will increase his usual rate by 10 kilometers per hour, it will take him one hour shorter to cover a distance of 200 kilometers. What is his usual speed in driving? 3. If you can row upstream to a landing 5 kilometers away and then row back to your starting point all in three hours and 20 minutes and if the river has an average current of 2 kilometers per hour, at what rate are you able to row in still water? 4. A horse travels 30 kilometers per hour faster than a mule. The horse goes 360 kilometers in two hours less time than the mule goes 360 kilometers. Find the speed of each animals? 5. An airplane flies 900 miles against a headwind of 25 miles per hour. The plane took 15 minutes longer for this flight than with a tailwind of 25 miles per hour. How fast could the plane fly in still air? 16

Lesson 4 Work Problems Work problems are important not only to students but also to industrialists,engineers, production executives and managers as well. This is becauseeveryone often meets work problems in real life.Common work problems have the following characteristics1. They ask you how long individual can finish a job alone, when working withothers when his rate of working is given; and2. They ask you how long a group of workers can finish a job, if rate of work of each member of the group is given.Examples:1. Mario takes 5 days more to do a job than Jose. Together they can do it in 6 days. How long does it take each alone to do the job?Solution:1st step: Identify the unknown. Jose No. of days to do job Amount of work done in 1 Mario x dayTogether 1 x+5 6 x 1 x+5 1 62nd step: Write the equation. 1+ 1 = 1 x x+5 6 17

3rd step: Operate the numbers to remove the parentheses grouping symbol. Multiply both sides by the least common denominator, 6x(x + 5). 6( x + 5 ) + 6x = x( x + 5 ) 6x + 30 + 6x = x2 + 5x4th step: Convert the equation to its standard form.Thus: 0 = x2 + 5x - 12x - 30 0 = x2 - 7x - 305th step: Factor the equation. 0 = ( x + 3 ) ( x - 10 )6th step: Solve for x.x+3 = 0 or x – 10 = 0 x = -3 x = 10The solutions of the fractional equation are 10 and -3. However, the solution of the problem is x = 10 and x stands for thenumber of days it takes Jose to do the work alone x + 5 = 15 is the number ofdays it takes Mario to do the work alone.2. tank has a supply pipe and an exhaust pipe. The exhaust pipe takes 5 minutes longer to empty the tank than for the supply pipe to fill it. If both are open, it takes the supply pipe 30 minutes to fill the tank. Find how long it takes the supply pipe to fill the tank when the exhaust pipe is closed.Solution:1st step: Identify the unknown.Let x = no. of minutes to fill in the tank 1 = part of the tank filled in one minute x 18

x + 5 = no. of minutes to empty the tank 1 = part of the tank emptied in one minute x+5 It takes 30 minutes to fill the tank with both pipes open2nd step: Write the equation. 30 - 30 = 1 x x+53rd step: Operate the numbers to remove the parentheses grouping symbol. Multiply both sides by the least common denominator which is x(x + 5). 30( x+ 5 ) - 30r = x( x + 5 ) 30x + 150 - 30x = x2 + 5x4th step: Convert the equation to its standard form.Thus: 0 = x2 + 5x - 1505th step: Factor the equation.0 = ( x + 15 ) ( x - 10 )6th step: Solve for x.x + 15 = 0 or x – 10 = 0x = -15 x = 10 It takes the supply pipe 10 minutes to fill the tank when the exhaust pipe isclosed. 19

Try this outAnswer the following problems: 1. Charlie can construct 100 identical boxes in two days less than Cholo. If they work together they can finish the work in 2 11 days. How long 12 will it take each of them to finish 100 boxes? 2. It takes Peter 6 hours longer than Luis to do a certain job. Together they can do it in 4 hours. How long would it take each working alone to do the job? 3. Mang Tano and his son Mario can finish planting their crops in 4 days if they work together. Working alone, Mario will need 2 1 days more than 3 the number of days that it will take his father to finish the work alone. How many days will each of the father and son need to finish the work if each will work alone? 4. Mr. Cruz can paint a house in 2 days less than his son. When they work together they can do the job in 4 4 days. How long would it take each 9 working alone to do job? Lesson 5 Problems Involving Quadratic Function (Maximum or Minimum Point) Quadratic functions have practical applications. Many applications of quadratic functions are in simple maximization and minimization problems. The technique for solving such problems is based on the fact that a quadratic function defined by f(x) = ax2 + bx + c attains its maximum value 20

( for a 〈 0 ) or minimum value ( for a 〉 0 ) at x = h = - b and the maximum / 2aminimum value is k = f(h) = 4ac − b2 . 4aExamples:1. The sum of two numbers is 24. Find the maximum product and the value of the two numbers.Solution: Let x = the first number 24 – x = the second number f(x) = the product of the two numbers y = f(x) = x(24 – x) = 24x – x2 or y = -x2 + 24x attains its maximum value at x = − b = − 24 = 12 2a 2(−1) The maximum product is 12(24 – 12) = 1442. A rectangular garden is x meters long and (18 – x) meters wide. If y is the area, find the maximum area that the garden could have. What could be its dimensions? Solution: Let x = the length 18 – x = width y(area) = x(18 – x) 21

= 18x – x2 or y = -x2 + 18x attains its maximum value at x = − b = −18 = 9 2a − 2 The maximum area is 9(18 – 9) = 18 square meters and the dimensions are length = 9 m and the width = 9 m.3. A rectangular lot is bounded on the side by a river and on the other three sides by a total of 80 m fencing. Find the possible dimensions of the lot. Solution: Step 1. The quantity to be maximized is the area of the lot. Step 2. Let x = be the width y = be the length If A is the area, then A = xy Step 3. Since the total length of fencing is 80 meters, 2x + y = 80 or y = 80 – 2x Substitute 80 – 2x for y in A = xy A = x(80 – 2x) = -2x2 + 80x Step 4. x = − b = − 80 = − 80 = 20 2a 2(−2) − 4 Area = -2x2 + 80x = -2(20)2 + 80(20) = -800 + 1600 = 800 Then maximum value of A is 800 and occurs when x = 20. Step 5. Since x = 20 and 2x + y = 80, y = 40. The dimensions of the largest possible rectangular lot are 40 meters by 20 meters. 22

4. A company charges P200 for each leather bag on order of 150 or less. The cost of each bag is reduced by P1 for each order in excess of 150. How many bags on order would result in a maximum revenue? what is the maximum revenue?Solution: Step 1. The quantity to be maximized is total revenue (R) in pesos from the leather bag to be sold. Step 2. Let x = be the number of bags ordered x - 150 = be number of bag ordered in excess of 150 200 – 1(x – 150) = 200 – x + 150 = 350 – x is the cost of each bag R = (cost per bag) (number of bags sold ) = (350 - x )x = 350x - x2 or -x2 + 350x Step 3. x = − b 2a = − 350 2(−1) = − 350 −2 = 175 R = -x2 + 350x = -( 175 )2 + 350( 175 ) = 30,625 The maximum value of R is 30,625 and it occurs when x = 175 23

Step 5. A purchase of 175 bags maximizes the company’s revenue. The maximum revenue is P30,625Try this outAnswer the following problems: 1. What are the dimensions of the largest rectangular field that can be enclosed with 60 meters of wire? 2. The sum of two numbers is 40. If one is x, what is the other? What is the maximum product that the 2 numbers could have? 3. A rectangular lot is bordered on one side by a string and on the other three sides by 600 meters of fencing. Find the dimensions of the lot if its area is a maximum. 4. A theater seats 2000 people and charges P10 for a ticket. At this price, all the tickets can be sold. A survey indicates that if the ticket price is increased by a peso, the number of tickets sold will decrease by 100. What ticket price result in the greatest revenue? Let’s summarize Steps in Solving word problems involving quadratic equation andfunction: 1. Identify your unknowns. 2. Write your equation. 3. Operate on the numbers. 4. Convert the equation into the standard form. Remember to change signs. 5. Factor the equation into its two corresponding linear equations. By factoring, you reduce the quadratic equation into two equivalent linear equations. 24

What have you learnedA. Define a variable and write a quadratic function to describe each of the following. 1. The product of two numbers whose sum is 40. 2. The product of two numbers whose difference is 25. 3. The area of a rectangle whose perimeter is 20 centimeter. 4. The product of two consecutive integers is nine less than the square of the second. 5. Write a quadratic function to describe the area of a circle in terms of its radius.B. Answer each of the following problems completely. 1. Find two consecutive even integers whose product is 168. 2. The sum of two numbers is 11. The difference of their squares is 11. What are the numbers? 3. The perimeter of a garden is 66 meters and its area is 270 sq. meters. Find the dimensions of the garden. 4. What two integers having a sum equal to 66 will have a maximum product? 5. Two bikers started at the same corner, one going east, the other going north. One biker is traveling at 3 kph faster than the other. After one hour, the two bikers are 15 kilometers apart. Find the rate of each. 25

Answer KeyHow much do you knowA. 1. b B. 1. A = s2 C. 1. 11 and 13 or -13 and -112. d 2. P = 2A 2. l = 16 cm. w = 6 cm3. c 3. 12x + 7 = 3x – 4 3. 9 and 134. b 4. c = 2π 4. 1445. d 5. d = rtTry this outLesson 11. x = 3 and x + 1 = 4 6. − 5 and 32. 14 and 16 35 7. 4 and 93. n = 9 8. -7 and -4 or 4 and 74. 5 and 15 9. 7 and 85. 5 and 11 10. n = 29Lesson 21. width = 4 meters and length = 8 meters2. length = 20meters and width = 18 meters3. base = 28 meters height = 21 meters4. The length of each side of the pool = 8 meters area of the Bermuda grass sidewalk = 132 square meters5. 1 m is the width of the walk 2 26

Lesson 31. B = 20kph, 2. The speed of Mr Cruz is 40 kph A = 24 kph3. The rate in still water 4. Rate of mule = 30 kph, = 4 km/hr Rate of horse = 60 kph5. r = 425 kphLesson 41. Charlie = 5 days, 2. Luis = 6 hours Cholo = 7 days Peter = 12 hours3. Mario = 9 1 days 4. son = 10 days 3 Mr. Cruz = 8 days Mang Tano = 7 daysLesson 51. width = 15 meters length = 15 meters2. 20 is the other number; 400 is the maximum product3. length = 300m width = 150 meters4. P25 ticket price would result to P22,500 greatest revenue.What have you learnedA. 1. y = x( 40 – x ) 2. y = x( x – 25 ) 5. A = πr2 3. A = l( 10 – l ) 4. P = x( x + 1 ) = ( x + 1 )2 - 9B. 1. -14 and -12 or 12 and 14 2. 6 and 5 3. length = 18 meters and width = 15 meters 4. the maximum product is 1089, the numbers are 33 and 33. 5. rate of slower biker = 9 kph rate of faster biker = 12 kph 27

Module 5 Circular Functions and Trigonometry What this module is about This module is about trigonometric equations and proving fundamentalidentities. The lessons in this module were presented in a very simple way so itwill be easy for you to understand solve problems without difficulty. Yourknowledge in previous lessons would be of help in the process What you are expected to learn This module is designed for you to: 1. state the fundamental identities 2. prove trigonometric identities 3. state and illustrate the sum and cosine formulas of cosine and sine 4. determine the sine and cosine of an angle using the sum and difference formulas. 5. solve simple trigonometric equationsHow much do you knowA. Answer the following:1. Which of the following does not equal to 1 for all A in each domain?a. sin2 A + cos2 A b. sec2 A - cos2 Ac. sin A sec A d. tan A cot A2. Simplify cos2 A sec A csc A

3. If sin ∝ = 12 and cos β = 4 , where ∝ and β are both in the first 13 5quadrant, find the values of cos (∝ + β ).4. Sec A is equal to c. 1 d. 1 . a. cos A b. sin A cos A sin A5. Express 1 − csc B in terms of cos B and Sin A. d. sin B −1 cot B cos B a. cos B – sinB b. 1 − sin B c. sin B – cos B cos B6. Simplify cosφ . sinφ cotφ a. 1 b. tanφ c. –csc φ d. -17. Multiply and simplify ( 1 – cos2 t ) ( 1 + tan2 t ).8. Express tan B ( sin B + cot B + cos B ) in terms of sec B.9. Compute sin 5π from the function of π and π . 12 4 610. Solve the equation cos A – 2sin A cos A = 0. 2

What you will do Lesson 1 Fundamental Trigonometric Identities To be able to simplify trigonometric expressions and solve trigonometricequations, you must be able to know the fundamental trigonometric identities.The Eight Fundamental Identities:A. Reciprocal Relations1. sec θ = 1 cosθ2. csc θ = 1 sinθ3. cot θ = 1 tanθB. Quotient Relations4. tan θ = sinθ cosθ5. cot θ = cosθ sinθC. Pythagorean Relations 6. cos2 θ + sin2 θ = 1 7. 1 + tan2 θ = sec2 θ 8. cot2 θ + 1 = csc2 θ 3

With the aid of this identities, you may now simplify trigonometricexpressions.Examples:Perform the indicated operation.a. ( 1 – sin x ) ( 1 + sin x )= 1 - sin2 x Product of sum & difference of 2 terms= cos2 x Since, cos2 θ + sin2 θ = 1, then 1 - sin2 x = cos2 xb. ( sec A – 1 ) ( sec A + 1 )= sec2 A - 1 Pythagorean Relation no. 2= tan2 Ac. tan θ ( cot θ + tan θ )= tan θ cot θ + tan2 θ= 1 + tan= sec2 θtan θ cot θ = 1, becausetan θ = sinθ cot θ = sinθ cosθ cosθtan θ cot θ = ( sinθ ) ( sinθ ) = 1 cosθ cosθd. cos x ( sec x - cos x ) = cos x sec x - cos2 x = 1 - cos2 x= sin2 xcos x sec x = 1, because sec x = 1 cos xtherefore, cos x sec x = cos x ( 1 ) = 1 cos x 4

e. cos B + sin2 B cos B, Least common denominator cos B Identity C. 6 Identity A. 1 = cos2 B + sin2 B cos B =1 cos B = sec BSimplify the following expressions to a single function. a. cos 2 A tan2 A = sin2 A cos 2 A ( sin2 A ) = sin2 A cos2 Ab. ( sin x + cos x )2 + ( sin x - cos x )2= sin2 x + 2sinx cos x + cos2x + sin2x - 2 sin x cos x + cos2x= sin2 x + cos2x + sin2x + cos2x=2since, cos2 θ + sin2 θ = 1c. cot B sec B sin B = 1 since, cot θ = cosθ and sec θ = 1 sinθ cosθ then, ( cosθ ) ( 1 ) sin θ = 1 sinθ cosθ 5

d. csc A - csc A cos2 A = sin A= csc A ( 1 - cos2 A ) Factor csc A= csc A ( sin2 A ) Identity C. 6= 1 ( sin2 A) Identity A. 2 Cancellation sinθ= sin Ae. cos3 B + cos B sin2 B= cos B ( cos2 B + sin2 B ) Factor cos B= cos B ( 1 ) Identity C. 6= cos B You are now ready to prove identities. In this lesson, you will prove thatone side of the equation is equal to the other side. You can work on either of thetwo sides to verify the expressions are equal or you can work on both equationsto arrive at an equal statement.Suggested Steps in Proving Identities 1. Start with the more complicated side and transform it into the simpler side. 2. Try algebraic operations such as multiplying, factorings, splitting single fractions and so on. 3. If other steps fail, express each function in terms of sine and cosine functions and then perform appropriate algebraic operations. 4. At each step, keep the other side of the identity in mind. This often reveals what one should do in order to get there. 6

Examples:a. Prove: cos B + tan B sin B = sec BSolution: Generally, we start with the more complicated side and transform it intothe other side using fundamental identities, algebra or other establishidentities.cos B + sin B ( sin B ) substituting sin B to tan B cos B cos Bcos B + sin2 B addition of fractions cos B 1 = sec B identity A. 1cos Bb. Prove that sec A - tan A sin A = cos ASolution: sec A - tan A sin A = cos A Substitute sin A to tan A sec A - tan A sin A = cos A 1 - sin A sin A =cos A cos A 1 - sin2 A = Subtraction of fraction cos A cos A 1 − sin2 A = Identity C. 6 cos A cos2 A = Division of fraction cos A cos A = cos A 7