# The Benefits of Using New Century Mathematics M2A Solutionrar for Learning and Teaching

## - Why is it important to learn? - What are the main topics covered in this book? H2: More about Trigonometric Functions - How to find the values of the six trigonometric ratios of any angle? - How to use the CAST rule to determine the quadrant of an angle? - How to solve trigonometric equations and inequalities? H2: Further Applications of Differentiation - How to find the stationary points and nature of a function? - How to sketch the graph of a function using differentiation? - How to apply differentiation to optimization problems? H2: Further Applications of Integration - How to find the area under a curve using integration? - How to find the volume of revolution using integration? - How to apply integration to kinematics problems? H2: Exponential and Logarithmic Functions - How to define and graph exponential and logarithmic functions? - How to use the laws of logarithms to simplify expressions? - How to solve exponential and logarithmic equations? H2: Binomial Theorem - How to expand binomial expressions using the binomial theorem? - How to use the binomial coefficients and Pascal's triangle? - How to apply the binomial theorem to probability problems? H1: Conclusion - Summarize the main points of the article - Emphasize the benefits of learning New Century Mathematics M2A - Provide some tips and resources for further study H2: FAQs - List five common questions and answers about New Century Mathematics M2A # Article with HTML formatting Introduction

New Century Mathematics M2A is a coursebook series written for senior secondary students who are taking Module 2 (Algebra and Calculus) of the extended part of the mathematics curriculum in Hong Kong. It consists of two volumes: M2A and M2B, covering topics such as trigonometric functions, differentiation, integration, exponential and logarithmic functions, binomial theorem, and more.

## New Century Mathematics M2A Solutionrar

Learning New Century Mathematics M2A is important for students who want to develop their mathematical skills and prepare for further studies or careers in science, engineering, or other fields that require advanced algebra and calculus. The coursebook series provides plenty of examples, exercises, and activities that help students master the concepts and techniques, as well as develop their problem-solving and critical thinking skills.

In this article, we will give you an overview of some of the main topics covered in New Century Mathematics M2A, and show you how to find solutions for some of the exercises. We will also provide some tips and resources for further study. Whether you are a student, a teacher, or a parent, we hope this article will help you learn more about New Century Mathematics M2A and appreciate its value.

## More about Trigonometric Functions

Trigonometric functions are functions that relate an angle to the ratios of the sides of a right-angled triangle. They are widely used in geometry, physics, engineering, and other fields that involve periodic phenomena or circular motion. In New Century Mathematics M2A, you will learn more about trigonometric functions, such as how to find their values, how to use them to solve equations and inequalities, and how to apply them to real-world problems.

### How to find the values of the six trigonometric ratios of any angle?

The six trigonometric ratios are sine (sin), cosine (cos), tangent (tan), cosecant (csc), secant (sec), and cotangent (cot). They are defined as follows:

sin Î¸ = opposite/hypotenuse

cos Î¸ = adjacent/hypotenuse

tan Î¸ = opposite/adjacent

csc Î¸ = hypotenuse/opposite

sec Î¸ = hypotenuse/adjacent

cot Î¸ = adjacent/opposite

To find the values of these ratios for any angle, you can use a calculator, a unit circle, or a reference triangle. A unit circle is a circle with radius 1 and center at the origin of a coordinate plane. A reference triangle is a right-angled triangle formed by dropping a perpendicular from a point on the unit circle to the x-axis. The angle Î¸ is measured from the positive x-axis to the terminal side of the angle.

For example, suppose you want to find the values of the six trigonometric ratios of 60. You can use a calculator and enter 60 in degree mode, then press the sin, cos, tan, csc, sec, and cot buttons. You should get the following results:

sin 60 0.866

cos 60 0.5

tan 60 1.732

csc 60 1.154

sec 60 2

cot 60 0.577

You can also use a unit circle and a reference triangle to find the values of the six trigonometric ratios of 60. First, draw a unit circle and locate the point (x, y) on the circle that corresponds to 60. You can use the formula x = cos Î¸ and y = sin Î¸ to find the coordinates of this point. For 60, you should get x = cos 60 = 0.5 and y = sin 60 = 3/2. Then, draw a perpendicular from this point to the x-axis and label the lengths of the sides of the reference triangle. You should get a triangle with hypotenuse 1, opposite side 3/2, and adjacent side 0.5. Finally, use the definitions of the six trigonometric ratios to find their values for 60. You should get the same results as above.

### How to use the CAST rule to determine the quadrant of an angle?

The CAST rule is a mnemonic device that helps you remember which trigonometric ratios are positive or negative in each quadrant of a coordinate plane. The word CAST stands for Cosine, All, Sine, and Tangent, which indicate the positive ratios in each quadrant.

To use the CAST rule, you can draw a square and divide it into four quadrants. Label each quadrant with one letter of CAST, starting from the bottom right and going counterclockwise. Then, for any angle Î¸, you can determine its quadrant by finding its reference angle Î±, which is the acute angle between the terminal side of Î¸ and the x-axis. The reference angle Î± is always positive and less than or equal to 90. The quadrant of Î¸ is then given by the letter of CAST that corresponds to Î±.

For example, suppose you want to find the quadrant of -135. First, find its reference angle Î± by adding or subtracting multiples of 360 until you get an angle between 0 and 360. For -135, you can add 360 and get 225. Then, find its reference angle Î± by subtracting or adding multiples of 90 until you get an acute angle. For 225, you can subtract 180 and get 45. Therefore, Î± = 45. Then, look at the CAST square and find which letter corresponds to 45. You should see that it is T for Tangent. Therefore, -135 is in the quadrant where tangent is positive, which is quadrant III.

### How to solve trigonometric equations and inequalities?

A trigonometric equation is an equation that involves one or more trigonometric functions of an unknown angle Î¸. A trigonometric inequality is an inequality that involves one or more trigonometric functions of an unknown angle Î¸.

To solve trigonometric equations and inequalities, you can use various methods such as algebraic manipulation, factoring, applying identities, using inverse functions, or using graphs. The general steps are as follows:

Rewrite the equation or inequality in terms of one trigonometric function if possible.

Isolate Î¸ on one side of the equation or inequality.

rad]). Use a calculator, a unit circle, or a reference triangle to find the exact or approximate values of Î¸.

Use the periodicity and symmetry of trigonometric functions to find all possible values of Î¸ in the given domain (usually [0, 360] or [0 rad, 2Ï€ rad]).

Check your solutions by substituting them back into the original equation or inequality.

For example, suppose you want to solve the equation sin Î¸ = 0.5 in the domain [0, 360]. You can follow these steps:

The equation is already in terms of one trigonometric function.

The angle Î¸ is already isolated on one side of the equation.

To find the values of Î¸ that satisfy the equation in one period, you can use a calculator and enter 0.5 in degree mode, then press the sin button. You should get Î¸ = 30. However, this is not the only solution, as there may be other angles that have the same sine value. To find the other solution, you can use the symmetry property of sine function, which states that sin (180 - Î¸) = sin Î¸. Therefore, if sin Î¸ = 0.5, then sin (180 - Î¸) = 0.5 as well. Solving for Î¸, you get Î¸ = 180 - 30 = 150. So, another solution is Î¸ = 150.

To find all possible values of Î¸ in the domain [0, 360], you can use the periodicity property of sine function, which states that sin (Î¸ + 360) = sin Î¸. Therefore, if sin Î¸ = 0.5, then sin (Î¸ + 360) = 0.5 as well. Adding 360 to both solutions, you get Î¸ = 30 + 360 = 390 and Î¸ = 150 + 360 = 510. However, these values are outside the domain [0, 360], so they are not valid solutions. Therefore, the only solutions in the domain [0, 360] are Î¸ = 30 and Î¸ = 150.

To check your solutions, you can substitute them back into the original equation and see if they make it true. For Î¸ = 30, you get sin 30 = 0.5, which is true. For Î¸ = 150, you get sin 150 = 0.5, which is also true. Therefore, both solutions are correct.

## Further Applications of Differentiation

Differentiation is a process that finds the rate of change of a function with respect to a variable. It is one of the most important tools in calculus and has many applications in mathematics and science. In New Century Mathematics M2A, you will learn more about differentiation, such as how to find the stationary points and nature of a function, how to sketch the graph of a function using differentiation, and how to apply differentiation to optimization problems.

### How to find the stationary points and nature of a function?

A stationary point of a function is a point where the derivative of the function is zero or undefined. A stationary point can be either a local maximum, a local minimum, or a point of inflection depending on how the function changes around that point. The nature of a stationary point can be determined by using the first derivative test or the second derivative test.

The first derivative test states that if f'(x) changes from positive to negative at x = c, then f(c) is a local maximum; if f'(x) changes from negative to positive at x = c, then f(c) is a local minimum; if f'(x) does not change sign at x = c, then f(c) is not a local extremum.

The second derivative test states that if f'(c) = 0 and f''(c) 0, then f(c) is a local minimum; if f'(c) = 0 and f''(c) = 0, then the test is inconclusive and the first derivative test should be used.

To find the stationary points and nature of a function, you can follow these steps:

Find the first derivative f'(x) of the function f(x) using the rules of differentiation.

Find the values of x that make f'(x) = 0 or undefined. These are the possible stationary points.

Use the first derivative test or the second derivative test to determine the nature of each stationary point.

For example, suppose you want to find the stationary points and nature of the function f(x) = x - 3x - 9x + 5. You can follow these steps:

The first derivative of f(x) is f'(x) = 3x - 6x - 9.

To find the values of x that make f'(x) = 0, you can factorize f'(x) as follows: f'(x) = 3x - 6x - 9 = 3(x - 2x - 3) = 3(x - 3)(x + 1). Therefore, f'(x) = 0 when x = 3 or x = -1. These are the possible stationary points.

To use the first derivative test, you can make a sign chart for f'(x) and see how it changes around x = 3 and x = -1. You should get something like this:

x ... -1 ... 3 ... --- --- --- --- --- --- f'(x) ... 0 ... 0 ... Sign + 0 - 0 + You can see that f'(x) changes from positive to negative at x = -1, so f(-1) is a local maximum. You can also see that f'(x) changes from negative to positive at x = 3, so f(3) is a local minimum.

To use the second derivative test, you need to find the second derivative f''(x) of the function f(x). The second derivative of f(x) is f''(x) = 6x - 6. Then, you need to evaluate f''(x) at x = -1 and x = 3. You should get f''(-1) = -12 and f''(3) = 12. Since f'(-1) = 0 and f''(-1) 0, f(3) is a local minimum by the second derivative test.

### How to sketch the graph of a function using differentiation?

Sketching the graph of a function is a useful way to visualize its behavior and properties. To sketch the graph of a function using differentiation, you can use the following steps:

Find the domain and range of the function.

Find the x- and y-intercepts of the function.

Find the first and second derivatives of the function.

Find the stationary points and nature of the function using the first and second derivative tests.

Find the intervals where the function is increasing or decreasing, and where it is concave up or concave down.

Find the asymptotes (horizontal, vertical, or oblique) of the function if any.

Plot the key points and features of the function on a coordinate plane and join them with a smooth curve.

For example, suppose you want to sketch the graph of the function f(x) = x - 3x - 9x + 5. You can follow these steps:

The domain of f(x) is all real numbers, and the range of f(x) is also all real numbers.

The x-intercepts of f(x) are the values of x that make f(x) = 0. To find them, you can factorize f(x) as follows: f(x) = x - 3x - 9x + 5 = (x - 5)(x + 1)(x + 1). Therefore, f(x) = 0 when x = 5 or x = -1. The x-intercepts are (5, 0) and (-1, 0). The y-intercept of f(x) is the value of f(0), which is 5. The y-intercept is (0, 5).

The first derivative of f(x) is f'(x) = 3x - 6x - 9, and the second derivative of f(x) is f''(x) = 6x - 6.

The stationary points and nature of f(x) are found in the previous section. They are (-1, 11), which is a local maximum, and (3, -22), which is a local minimum.

The intervals where f(x) is increasing or decreasing can be found by using the sign chart for f'(x), which was also shown in the previous section. You can see that f'(x) > 0 when x 3, and f'(x) 3, and decreasing when -1

x ... 1 ... --- --- --- --- f''(x) ... 0 ... Sign - 0 + You can see that f''(x) 0 when x > 1. Therefore, f(x) is concave down when x 1.

The asymptotes of f(x) are lines that the graph of f(x) approaches as x or y gets very large or very small. There are three types of asymptotes: horizontal, vertical, and oblique. For this function, there are no asymptotes because it is a polynomial function and its degree is odd. Therefore, as x gets very large or very small, so does y.

To plot the key points and features of f(x) on a coordinate plane, you can use a graphing calculator or software, or draw it by hand. You should mark the x- and y-intercepts, the stationary points, the inflection point, and the intervals where f(x) is increasing or decreasing, and where it is concave up or concave down. You should also label the axes and the scale. You should get something like this:

## ![Graph of f(x)](https://i.imgur.com/9w1Y8XH.png) Further Applications of Integration

Integration is a process that finds the total change of a function with respect to a variable. It is the inverse of differentiation and has many applications in mathematics and science. In New Century Mathematics M2A, you will learn more about integration, such as how to find the area under a curve using integration, how to find the volume of revolution using integration, and how to apply integration to kinematics problems.

### How to find the area under a curve using integration?

The area under a curve is the region bounded by the curve, the x-axis, and two vertical lines. To find the area under a curve using integration, you can use the following steps:

Find the points of intersection of the curve and the x-axis. These are the limits of integration.

Write an integral expression for the area under the curve. The integrand is the function that defines the curve, and the variable of integration is x.

Evaluate the integral using the fundamental theorem of calculus or other methods.

For example, suppose you want to find the area under the curve f(x) = x - 3x - 9x + 5 from x = -1 to x = 5. You can follow these steps:

The points of intersection of f(x) and the x-axis are found by solving f(x) = 0. This was done in a previous section and the solutions are x = -1 and x = 5. These are the limits of integration.

The integral expression for the area under f(x) from x = -1 to x = 5 is A = -1 (x - 3x - 9x + 5) dx.

To evaluate the integral, you can use the fundamental theorem of calculus, which states that if F(x) is an antiderivative of f(x), then a f(x) dx = F(b) - F(a). To find an antiderivative of f(x), you can use the rules of integration. For f(x) = x - 3x - 9x + 5, you can use the power rule and get F(x) = x/4 - x - 9x/2 + 5x + C, where C is an arbitrary constant. Then, you can plug in x = -1 and x = 5 into F(x) and subtract them. You should get A = F(5) - F(-1) = (625/4 - 125 - 225/2 + 25) - (1/4 + 1 + 9/2 - 5) = (125/4 - 101/2) - (-17/4 + 11/2) = (142/4) - (-6/4) = 148/4 = 37. Therefore, the area under f(x) from x = -1 to x = 5 is 37 square units.

### How to find the volume of revolution using integration?

The volume of revolution is the volume of a solid obtained by rotating a plane region around an axis. To find the volume of revolution using integration, you can use one of two methods: the disk method or the shell method. The disk method is used when the region is bounded by a curve and a line parallel to the axis of rotation. The shell method is used when the region is bounded by two curves and two lines perpendicular to the axis of rotation.

The disk method states that if R is a region bounded by y = f(x), y = g(x), x = a, and x = b, where f(x) g(x), and if R is rotated around the x-axis, then the volume of revolution is V = a Ï€[f(x)] dx.

The shell method states that if R is a region bounded by y = f(x), y = g(x), x = a, and x = b, where f(x) g(x), and if R is rotated around the y-axis, then the volume of revolution is V = a 2Ï€x[f(x) - g(x)] dx.

For example, suppose you want to find the volume of revolution of the region bounded by y = x - 3x - 9x + 5, y = 0, x = -1, and x = 5, when it is rotated around the x-axis. You can use the disk method as follows:

The region R is bounded by y = f(x) = x - 3x - 9x + 5, y = g(x) = 0, x = a = -1, and x = b = 5. The function f(x) is always greater than or equal to g(x) in this region.

The volume of revolution is V = -1 Ï€[f(x)] dx