GENAM301_Basic algebra and trigonometry

[Virtual Presenter] This module covers the fundamental concepts of basic algebra and trigonometry. Ten credits have been allocated to this module, highlighting its significance in our curriculum. The learning hours required for this module total 100, implying a substantial commitment of time and effort from students. This module encompasses all relevant sectors and sub-sectors, making it applicable across various fields and industries. As outlined in the purpose statement, the goal of this module is to provide learners with essential knowledge in applying basic algebra and trigonometry, thereby enhancing their ability to succeed in their engineering careers..

[Audio] The elements of competence and performance criteria are outlined below. These include learning units such as solving linear equations and inequalities algebraically or graphically, discussing parametric equations, solving simultaneous linear equations, solving quadratic equations, and solving quadratic inequalities. Additionally, appropriate descriptions of angles based on rotating an initial side about a fixed point are also included..

[Audio] Trigonometric ratios can be determined appropriately based on isosceles right-angled triangles and equilateral triangles. Trigonometric identities can be described by comparing the trigonometric ratios of two defined angles. Trigonometric equations can be solved perfectly using trigononic ratios. A given triangle can also be solved based on the Pythagorean theorem and trigonometric ratios..

[Audio] The algebraic method can be used to solve a linear equation. First, we must isolate the variable x on one side of the equation. The given equation is 9x + 7 = 4x - 16. Adding 16 to both sides gives 9x + 7 + 16 = 4x - 16 + 16, which simplifies to 9x + 23 = 4x. Subtracting 4x from both sides results in 9x - 4x + 23 = 4x - 4x, which simplifies to 5x + 23 = 0. Subtracting 23 from both sides gives 5x + 23 - 23 = 0 - 23, resulting in 5x = -23. Dividing both sides by 5 yields x = -23/5..

[Audio] When solving a linear inequality, we use similar methods to those used when solving a linear equation, but with some key differences. When multiplying an inequality by a negative real number, the direction of the inequality symbol will change. If we have an inequality like 9x + 7 ≥ 4, and we multiply both sides by -1, the inequality would become -9x - 7 ≤ 4. When interchanging the right-hand side and the left-hand side of an inequality, the direction of the inequality symbol will also change. If we have an inequality like 9x + 7 ≥ 4, and we rearrange it to get 4 ≥ 9x + 7, the inequality symbol would need to be flipped to get 4 ≤ 9x + 7. To solve the inequality 9x + 7 ≥ 4, we first isolate the variable term on one side of the inequality. We can subtract 7 from both sides to get 9x ≥ -3. Then, we divide both sides by 9 to get x ≥ -3/9 or x ≥ -1/3. To solve the inequality 9x + 7 ≤ 4, again we start by isolating the variable term on one side of the inequality. Subtracting 7 from both sides gives us 9x ≤ -3. Dividing both sides by 9 results in x ≤ -3/9 or x ≤ -1/3. The solution to this inequality is the same as the previous one, but with a less-than-or-equal-to symbol instead of a greater-than-or-equal-to symbol. This makes sense because the original inequality was asking for values of x that are less than or equal to -1/3, rather than strictly greater than -1/3. By understanding how to handle these special cases when solving linear inequalities, we can ensure that our solutions accurately reflect the relationships between the variables and constants involved..

[Audio] A parameter is a variable used to describe a set of related values, while a parametric equation is a mathematical equation that involves one or more parameters. Certain coefficients may contain letter variables, making them parametric. When solving and discussing the equation for parameters only, specific steps must be followed. The equation should be approached by focusing solely on the parameters..

[Audio] The coefficients a, b, and c in a parametric quadratic equation are related to a real parameter which may not be fully determined. The root of the parametric quadratic equation depends on the specific values attributed to that parameter..

[Audio] Simultaneous linear equations are equations involving two variables, where each equation is linear. We can solve these equations by either algebraic methods or graphical methods. Algebraic methods involve using techniques such as substitution or elimination to find the values of the variables. Graphical methods involve plotting the lines represented by the equations on a coordinate plane and finding the point of intersection..

[Audio] The substitution method involves expressing one variable in terms of the other. This is done by isolating one variable in one of the equations. Once this is done, we can substitute this expression into one of the original equations to solve for the remaining variable. The elimination method is used when neither variable is given as the subject of another. We can add or subtract the equations to eliminate one of the variables, making it easier to solve for the other variable. The comparison method is not typically used in practice, but it's worth mentioning as an option. Cramer's rule is a more advanced method that involves calculating determinants to solve the system of equations. Each of these methods has its own strengths and weaknesses, and the choice of which one to use will depend on the specific problem being solved. By understanding these different approaches, we can better tackle complex systems of linear equations and arrive at accurate solutions..

[Audio] To solve this system of linear equations, we can express x in terms of y using both equations. From equation (i) 3x - 2y = 2, we can add 2y to both sides to get 3x = 2 + 2y. Then, dividing both sides by 3 gives us x = (2 + 2y)/3. This simplifies to x = (2y + 2)/3, which is our first expression for x in terms of y. Now, looking at equation (ii) 7x + 3y = 43, if we subtract 3y from both sides, we get 7x = 43 - 3y. Dividing both sides by 7 yields x = (43 - 3y)/7. Therefore, we now have two expressions for x in terms of y: x = (2y + 2)/3 and x = (-3y + 43)/7. To proceed with the comparison method, we'll set these two expressions equal to each other, resulting in the equation (2y + 2)/3 = (-3y + 43)/7. We can simplify this equation to solve for y. Multiplying both sides by 21 to clear the fractions gives us 7(2y + 2) = 3(-3y + 43). Expanding both sides results in 14y + 14 = -9y + 129. Adding 9y to both sides gives us 23y + 14 = 129. Subtracting 14 from both sides yields 23y = 115. Finally, dividing both sides by 23 gives us y = 5. With the value of y known, we can substitute it back into one of the original equations to solve for x. Let's use equation (iii): x = (2y + 2)/3. Substituting y = 5 gives us x = (2*5 + 2)/3, which simplifies to x = (10 + 2)/3. Further simplification yields x = 12/3, and finally, x = 4. Therefore, the solution to the system of linear equations is x = 4 and y = 5..

[Audio] Multiplying both sides by 21, we get 7(2y + 2) = 3(-3y + 43). Expanding both sides gives us 14y + 14 = -9y + 129. Now, let's isolate the variable y by moving all terms involving y to one side of the equation. Subtracting 14y from both sides gives us 14 = -9y + 129. Next, subtracting 129 from both sides yields -115 = -9y. To solve for y, we divide both sides by -9, giving us y = 115/23. Therefore, y equals 5. Now that we have found the value of y, we can substitute it into either equation (iii) or equation (iv) to find the value of x. Let's use equation (iii). Substituting y = 5 into the equation gives us x = (2 * 5 + 2)/3. Simplifying the expression inside the parentheses first, we get x = (10 + 2)/3. Combining the terms inside the parentheses yields x = 12/3. Finally, dividing 12 by 3 gives us x = 4. So, the values of x and y are 4 and 5, respectively. This method of solving simultaneous equations is called the comparison method. By comparing the values of x obtained from equation (i) and equation (ii), we can form an equation in y. Similarly, comparing the two values of y allows us to form an equation in x..

[Audio] Graphical methods for solving two simultaneous linear equations involve plotting the lines representing each equation on a coordinate plane and identifying the point of intersection, where the two lines meet. This point represents the solution to the system of equations..

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[Audio] A quadratic equation is an expression in the form ax^2 + bx + c = 0, where a, b, and c are constants, and x is the variable..

[Audio] The square root property can be used to solve a quadratic equation in the form of ax squared equals, plus bx, plus c, where a, b, and c are constants. If A equals B squared, then the square root of A equals plus minus the square root of B. To apply this property, we must transform the equation into the form of a perfect square on one side and a constant on the other side. Taking the square root of each side gives us both positive and negative values. However, if we are dealing with a negative number under the square root sign, we will not get any real solutions, and we may need to consider using imaginary numbers instead. The equation x squared, plus 7x, plus 10 equals zero can be solved by applying the square root property. Rearranging the terms and completing the square allows us to write the left-hand side of the equation as a perfect square. Applying the square root property to the equation yields the solutions plus minus 5i. Therefore, the solutions to the equation are the square roots of the constant term, which are plus minus 5i..

[Audio] The quadratic formula is used when it's not possible to factorise or use the square root property to solve an equation. The quadratic formula is given by x = (-b ± √(b² - 4ac)) / 2a. In this case, a = 1, b = 8, and c = -2. Plugging these values into the formula, we get x = (-(8) ± √((8)² - 4(1)(-2))) / 2(1). Simplifying further, we get x = (-8 ± √(64 + 8)) / 2, which becomes x = (-8 ± √72) / 2. We can simplify √72 to √(36*2), which equals 6√2. So now we have x = (-8 ± 6√2) / 2. To finish solving the equation, we need to isolate x, which means getting rid of the fraction. We do this by multiplying both sides of the equation by 2. This gives us 2x = -8 ± 6√2. Now we can divide both sides by 2 to finally get x = (-8 ± 6√2) / 2. Therefore, the solutions to the equation are x = (-8 + 6√2) / 2 and x = (-8 - 6√2) / 2..

[Audio] To solve these exercises, we need to apply the concepts learned so far in this module. Key points covered in the previous slides should be reviewed. These concepts will then be used to work through each exercise step by step. The goal is to ensure that we understand the underlying principles and can confidently apply them to real-world problems. Let's dive into the first exercise and see how we can put our knowledge into practice..

[Audio] The graphical resolution of a quadratic equation involves constructing a parabola. This can be done by plotting points on a coordinate plane using the equation's coefficients. By examining the shape and position of the parabola, we can determine the solutions to the equation..

[Audio] The y-intercept is one of the key points that should be included in the graph when graphing a function. The y-intercept is the location where the graph crosses the y-axis. It can be found by setting x equal to zero and solving for y. Another important point is the x-intercept, which is the location where the graph crosses the x-axis. To find this point, set y equal to zero and solve for x. The vertex of the graph is a critical point that determines whether the graph has a minimum or maximum value. It is located at the lowest or highest point on the graph. The line of symmetry, also known as the axis of symmetry, is a vertical line that passes through the vertex and divides the graph into two symmetrical parts. This line is essential in understanding the shape and behavior of the graph. By including these special points in the graph, we can gain a deeper understanding of the function's properties and behavior..

[Audio] When determining the solution of a quadratic equation, we use the graphical method. We start by drawing a graph of the quadratic function using a table of values. By examining the graph, we can identify the x-coordinates of the points where the curve intersects the horizontal axis. These x-coordinates represent the solutions or roots of the equation. There are three possible scenarios: first, if the graph cuts the x-axis at two distinct points, the quadratic has two distinct roots; second, if the x-axis is a tangent to the curve, the two roots are equal, resulting in a single solution; third, if the curve does not intersect or touch the x-axis, there are no real solutions..

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[Audio] The two main methods available to us when we want to solve a quadratic inequality are factoring and graphical resolution. Factoring involves breaking down the given inequality into simpler expressions, determining its roots, studying the sign of the expression between these roots, and finally determining the intervals of solutions. To do this, we first factorize the given inequality. Then, we determine its roots by setting each factor equal to zero and solving for x. Next, we choose test points within each interval created by the roots and study the sign of the expression. Finally, we determine the intervals of solutions based on the sign of the expression. Graphical resolution involves plotting a parabola and shading the region satisfying the given inequality. By doing so, we can visually identify the intervals of solutions. For instance, if we're given the quadratic inequality 0 > x^2 - 3x + 4, we can solve it graphically by plotting the parabola y = x^2 - 3x + 4 and shading the region above the curve. From this graph, we can see that the solution to the inequality is the interval (-∞, 1) ∪ (4, ∞)..

[Audio] The square root property can be used to solve certain types of quadratic equations. This property states that if we have an equation of the form ax^2 + bx + c = 0, where a, b, and c are constants, and if we can rewrite it in the form (x + d)^2 = e, where d and e are also constants, then we can use the fact that the square root of a number is equal to either its positive or negative square root. In other words, if we have an equation like this, we can take the square root of both sides and get two possible solutions: x + d = ±√e. We can then simplify these expressions to get our final answers. However, not all quadratic equations can be solved using the square root property. We need to check whether the equation can be rewritten in the desired form before applying this method..

[Audio] The angle is a measure of the space between two rays with a common endpoint. It is typically measured by the amount of rotation required to get from its initial side to its terminal side. The initial side is where the rotational measure begins, while the terminal side is where it ends. The vertex is the point where these two rays meet, serving as the common endpoint of the angle. When rotating an angle, we are essentially turning its terminal side around the origin, assuming it is in standard position..

[Audio] Angles can be measured in different units. We have been using degrees as one of these units so far. However, there's another way of measuring angles that's often more convenient, known as radians. To understand what a radian is, let's consider how it's defined. One radian is the measure of the angle formed by wrapping the radius of a circle along its exterior. This concept is crucial for various mathematical and scientific applications. Now, let's move on to discussing units conversion..

[Audio] The Pythagorean theorem is a fundamental concept in geometry that describes the relationship between the lengths of the sides of a right-angled triangle. It states that the square of the length of the hypotenuse, the side opposite the right angle, is equal to the sum of the squares of the lengths of the other two sides. In mathematical terms, this can be expressed as c^2 = a^2 + b^2, where c is the length of the hypotenuse, and a and b are the lengths of the other two sides. This theorem has numerous applications in various fields, including physics, engineering, and architecture..

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[Audio] Trigonometric ratios are used to describe the relationships between the sides and angles of triangles. They are calculated using the sine, cosine, and tangent functions. In this unit, we will focus on determining these ratios for different types of triangles, including right-angled triangles and equilateral triangles..

[Audio] Trigonometric ratios can be calculated for special angles such as zero degrees, thirty degrees, forty-five degrees, and sixty degrees. These calculations involve determining the sine, cosine, and tangent values for these specific angles..

[Audio] Trigonometric identities enable us to relate the trigonometric ratios of different angles. Complementary angles are those whose sum is ninety degrees. Thirty degrees and sixty degrees are examples of complementary angles. The sine of one angle equals the cosine of the other. Supplementary angles are those whose sum is one hundred and eighty degrees. Forty-five degrees and one hundred and thirty-five degrees are examples of supplementary angles. The trigonometric ratios of supplementary angles are related but in a different manner. Trigonometric ratios of the sum or difference of two angles can be simplified using the formulas for the sine, cosine, and tangent of the sum or difference of two angles. These formulas help us simplify expressions involving these trigonometric functions. Trigonometric ratios of double angles can be simplified using the formulas for the sine, cosine, and tangent of twice an angle. These formulas help us simplify expressions involving these trigonometric functions. Applying these trigonometric identities enables us to solve problems involving trigonometric functions more efficiently..

[Audio] When solving trigonometric equations, we need to consider various forms that these equations can take. One such form is represented by the equation sin(kx + b) = sin(a). We can simplify this equation using the identity sin(A) = sin(B), which implies A = B + 2nπ or A = 2mπ - B, where n and m are integers. Applying this identity to our equation, we get kx + b = a + 2nπ or kx + b = 2mπ - a. From here, we can solve for x by isolating it on one side of the equation. The equation sin(4/3)x + sin(3/4)x = 0 can be rewritten as sin(4/3)x = -sin(3/4)x. Using the identity mentioned earlier, we can simplify this equation to find the values of x that satisfy it. By applying the appropriate formulas and simplifying the resulting expressions, we can determine the solutions to the equation..

[Audio] The sine, cosine and tangent functions can be used to solve trigonometric equations. These functions relate the ratio of the lengths of the sides of a right-angled triangle. They are used to find the unknown angle when the lengths of the other two sides are known. The sine function relates the length of the opposite side to the hypotenuse, the cosine function relates the length of the adjacent side to the hypotenuse, and the tangent function relates the length of the opposite side to the adjacent side. By using these relationships, we can solve trigonometric equations such as sin(x) = 0.5, cos(x) = 0.8, and tan(x) = 0.6. We can also use the inverse functions to find the value of x. For example, if we have the equation sin(x) = 0.5, we can use the inverse sine function to find the value of x. Similarly, if we have the equation cos(x) = 0.8, we can use the inverse cosine function to find the value of x. And if we have the equation tan(x) = 0.6, we can use the inverse tangent function to find the value of x. These inverse functions allow us to find the value of x by reversing the operation of the original function. In this case, the inverse sine function returns the angle whose sine is equal to 0.5, the inverse cosine function returns the angle whose cosine is equal to 0.8, and the inverse tangent function returns the angle whose tangent is equal to 0.6. By using these inverse functions, we can solve trigonometric equations and find the value of x..

[Audio] Solving the equation asinx + bcosx = c involves using trigonometric identities to simplify the expression and then applying algebraic techniques to isolate the variable x. The identity sin^2(x) + cos^2(x) = 1 can be used to rewrite the equation in terms of sine and cosine functions only. Sum-to-product identities can also be used to combine the sine and cosine terms into a single product term. After simplification, standard algebraic methods such as factoring or substitution can be used to solve for x..

[Audio] To solve this equation, let's first simplify it by using the identity cos(2θ) = 1 - 2sin^2(θ). Substituting this into the equation, we get 0 = 3cos(2θ) + 2x. Now, we can use the double-angle formula for cosine, which states that cos(2θ) = 1 - 2sin^2(θ). Substituting this into our equation, we get 0 = 3(1 - 2sin^2(θ)) + 2x. Expanding and simplifying, we get 0 = 3 - 6sin^2(θ) + 2x. Rearranging terms, we get 6sin^2(θ) = 3 + 2x. Dividing both sides by 6, we get sin^2(θ) = (3 + 2x)/6. Taking the square root of both sides, we get sin(θ) = ±√((3 + 2x)/6). Since we're dealing with a triangle, we'll consider only the positive value of sin(θ), so sin(θ) = √((3 + 2x)/6). Now, we can use the fact that sin(θ) = opposite side / hypotenuse to relate θ to the sides of the triangle. However, since we're not given any specific values for the sides, we cannot proceed further without additional information. Therefore, we need more data about the triangle to apply the sine law effectively..

[Audio] The Cosine Law helps us solve problems involving the lengths of sides and the measures of angles when working with triangles. The law applies when we're given either two sides and a contained angle, or when we're given all three sides and need to find one or more of the angles. To apply the Cosine Law, we use the formula: a^2 + b^2 - 2ab * cos(C) = c^2, where a, b, and c are the lengths of the sides, and C is the measure of the angle between sides a and b. This formula allows us to find the length of a side, or the measure of an angle, depending on what's given. If we're given two sides and a contained angle, we can rearrange the formula to solve for the length of the third side. When we're given all three sides, we can use the formula to find the measure of one or more of the angles. By applying the Cosine Law, we can effectively solve various types of triangle problems..

[Audio] Complex numbers are a fundamental concept in mathematics with numerous applications in various fields including engineering. A complex number is a number that can be expressed in the form a + bi, where a and b are real numbers and i is the imaginary unit which satisfies the property i^2 = -1. The real part of a complex number is represented by a, while the imaginary part is represented by bi. The set of all complex numbers is denoted by C. Complex numbers can be represented geometrically using an Argand diagram. An Argand diagram is a graphical representation of complex numbers where each point in the plane corresponds to a unique complex number. This representation allows us to visualize and manipulate complex numbers more effectively..

[Audio] Complex numbers can be added by adding their real parts and imaginary parts separately. For example, if we have two complex numbers 3 + 4i and 2 + 5i, we add them by adding their real parts (3 + 2) and their imaginary parts (4i + 5i), resulting in the sum 5 + 9i. Similarly, complex numbers can be subtracted by subtracting their real parts and imaginary parts separately. Multiplication of complex numbers involves multiplying the real part of one number with the real part of another, the real part of one number with the imaginary part of another, the imaginary part of one number with the real part of another, and the imaginary part of one number with the imaginary part of another, and then combining these products according to specific rules. Division of complex numbers also follows specific rules..

[Audio] To add these complex numbers, we need to add the real parts together and the imaginary parts together separately. So, (3 + 4i) + (2 - 7i) = (3 + 2) + (4i - 7i). Now, let's simplify this expression by adding the real parts and the imaginary parts separately. We get 5 - 3i. Therefore, the sum of the complex numbers 3 + 4i and 2 - 7i is 5 - 3i..

[Audio] The modulus of a complex number is a measure of its distance from the origin in the complex plane. It can be calculated using the formula |z| = √(a² + b²), where z = a + bi is a complex number with real part a and imaginary part b..