Hello students, welcome to today's mathematics class. I'm so happy to see you all here, ready to learn something new and exciting. Today, we are going to begin a very interesting chapter, Chapter 8, which is Introduction to Trigonometry. Now, I know some of you might have heard this word before, and some of you might be wondering what on earth trigonometry is. Don't worry, by the end of this lesson, you will not only understand what it means but also be able to solve problems related to it. So let's begin together.
Students, have you ever looked at the Qutub Minar in Delhi and wondered how tall it is? Or have you ever tried to imagine how wide a river is, without actually measuring it? Or think about a hot air balloon in the sky - can we find out how high it is from the ground? Well, all these questions and many more can be answered using a branch of mathematics called trigonometry. The word trigonometry comes from Greek words - 'tri' meaning three, 'gon' meaning sides, and 'metron' meaning measure. So essentially, trigonometry is the study of relationships between the sides and angles of a triangle. Isn't that wonderful? This concept was first used by ancient astronomers to find out distances of stars and planets from the Earth, and even today, it forms the basis of many engineering and physical science calculations.
Now, let's understand what a trigonometric ratio is. Consider a right triangle ABC, where angle B is 90 degrees. Let me draw this in your mind - we have triangle ABC with right angle at B. Now, consider angle A, which is an acute angle. Now, look at the side BC - this side is opposite to angle A, meaning it faces angle A. The side AC is the hypotenuse, which is the longest side opposite the right angle. And side AB is adjacent to angle A, meaning it forms part of the angle A. Now, students, this is very important - when we change our perspective and look at angle C instead of angle A, the positions of these sides change completely. For angle C, the side AB becomes opposite, and BC becomes adjacent. So always pay attention to which angle you are considering.
Now, we define certain ratios involving these sides, and these are called trigonometric ratios. For angle A in our right triangle ABC, we define:
Sine of angle A, written as sin A, is the ratio of the side opposite to angle A divided by the hypotenuse. So sin A equals BC divided by AC.
Cosine of angle A, written as cos A, is the ratio of the side adjacent to angle A divided by the hypotenuse. So cos A equals AB divided by AC.
Tangent of angle A, written as tan A, is the ratio of the side opposite to angle A divided by the side adjacent to angle A. So tan A equals BC divided by AB.
Now, students, there are three more ratios which are the reciprocals of these three. Cosecant of angle A, written as cosec A, is the reciprocal of sine, so it equals AC divided by BC. Secant of angle A, written as sec A, is the reciprocal of cosine, so it equals AC divided by AB. And cotangent of angle A, written as cot A, is the reciprocal of tangent, so it equals AB divided by BC.
Let me write these down once more so they are clear in your mind. Sin A equals opposite over hypotenuse. Cos A equals adjacent over hypotenuse. Tan A equals opposite over adjacent. And their reciprocals are cosec A, sec A, and cot A respectively. Now, here's an interesting relationship - tan A can also be written as sin A divided by cos A. Similarly, cot A equals cos A divided by sin A. Keep these relationships in mind as we will use them frequently.
Now, students, I want to tell you something fascinating about the history of these terms. The first use of the word 'sine' in the way we use it today was in the work Aryabhatiyam by the great Indian mathematician Aryabhata, around 500 AD. Aryabhata used the word 'ardha-jya' for half-chord, which was shortened to 'jya' or 'jiva'. When this was translated into Arabic, it became 'jiva', and when the Arabic version was translated into Latin, it became 'sinus', which means curve. Eventually, it became 'sine'. The terms cosine and tangent came much later. The cosine function arose from computing the sine of the complementary angle, and Aryabhata called it 'kotijya'. Isn't it wonderful that Indian mathematics has contributed so much to trigonometry?
Now, here's an important observation. If we take any point on the hypotenuse or its extension, and draw a perpendicular to the base, we get a similar triangle. And here's the key point - the trigonometric ratios of an angle remain the same regardless of the size of the triangle, as long as the angle itself doesn't change. This is because the triangles are similar, and corresponding sides are proportional. So, students, remember this - the values of trigonometric ratios depend only on the angle, not on the size of the triangle.
Now, let's look at an example. Suppose in a right triangle ABC, we know that sin A equals 1/3. This means BC/AC equals 1/3, so the sides BC and AC are in the ratio 1:3. Let BC equal k, then AC equals 3k. Now we need to find AB. Using Pythagoras theorem, AB squared equals AC squared minus BC squared, which is (3k) squared minus (k) squared, which equals 9k squared minus k squared, which is 8k squared, which is (2√2 k) squared. So AB equals 2√2 k. Now we can find cos A, which is AB/AC, and that equals 2√2 k divided by 3k, which is 2√2/3. Similarly, we can find all other trigonometric ratios.
Now, here's an important note - since the hypotenuse is the longest side in a right triangle, the value of sin A or cos A is always less than 1, though it can equal 1 in special cases.
Let me now show you how to find all trigonometric ratios when one of them is given. Consider Example 1 from your textbook. We are given that tan A equals 4/3, and we need to find the other trigonometric ratios. Let's draw a right triangle where tan A equals 4/3. Since tan A equals opposite over adjacent, let BC equal 4k and AB equal 3k, where k is any positive number. Now, using Pythagoras theorem, AC squared equals AB squared plus BC squared, which is (3k) squared plus (4k) squared, which is 9k squared plus 16k squared, which is 25k squared. So AC equals 5k. Now we can find all the ratios. Sin A equals opposite over hypotenuse, which is BC/AC, which is 4k/5k, which is 4/5. Cos A equals adjacent over hypotenuse, which is AB/AC, which is 3k/5k, which is 3/5. Now, cot A is the reciprocal of tan A, so it equals 3/4. Cosec A is the reciprocal of sin A, so it equals 5/4. And sec A is the reciprocal of cos A, so it equals 5/3. See how easy it is? Once you know one ratio, you can find all the others.
Now, let's look at another example. In Example 3, we have a right triangle ABC, right-angled at C, where AB equals 29 units, BC equals 21 units, and angle ABC is theta. We need to find cos squared theta plus sin squared theta, and cos squared theta minus sin squared theta. First, we find AC using Pythagoras theorem. AC squared equals AB squared minus BC squared, which is 29 squared minus 21 squared, which is 841 minus 441, which is 400, so AC equals 20 units. Now, sin theta equals opposite over hypotenuse, which is AC/AB, which is 20/29. Cos theta equals adjacent over hypotenuse, which is BC/AB, which is 21/29. Now, cos squared theta plus sin squared theta equals (20/29) squared plus (21/29) squared, which is (400 + 441)/841, which is 841/841, which equals 1. This is a very important result - for any angle, sin squared theta plus cos squared theta always equals 1. We'll discuss this more when we come to trigonometric identities. And cos squared theta minus sin squared theta equals (21/29) squared minus (20/29) squared, which is (441 - 400)/841, which is 41/841.
Now, students, I want you to remember that sin squared A means (sin A) squared. We write it as sin²A for convenience. But be careful - cosec A is not the same as sin⁻¹A. The notation sin⁻¹A means the inverse sine, which is a different concept that you will study in higher classes.
Now, let's move on to a very important section - finding trigonometric ratios for specific angles like 0°, 30°, 45°, 60°, and 90°. These are very important and you should memorize them.
Let's start with 45°. Consider a right triangle where one angle is 45°. Then the other angle must also be 45° because the sum of angles in a triangle is 180°, and one angle is 90°. So in a right triangle with angles 45°, 45°, and 90°, the two legs are equal. Let each leg be of length a. Then by Pythagoras theorem, the hypotenuse squared equals a squared plus a squared, which is 2a squared, so the hypotenuse equals a√2. Now, sin 45° equals opposite over hypotenuse, which is a/(a√2), which is 1/√2. Cos 45° equals adjacent over hypotenuse, which is also 1/√2. Tan 45° equals opposite over adjacent, which is a/a, which is 1. The reciprocals are cosec 45° equals √2, sec 45° equals √2, and cot 45° equals 1.
Now, let's find the ratios for 30° and 60°. Consider an equilateral triangle ABC, where each angle is 60°. Draw a perpendicular from A to BC, calling it AD. Now, triangle ABD is a right triangle with angle BAD equal to 30° and angle ABD equal to 60°. Since the triangles ABD and ACD are congruent, BD equals half of BC. Let AB equal 2a. Then BD equals a. Using Pythagoras theorem, AD squared equals AB squared minus BD squared, which is (2a) squared minus a squared, which is 4a squared minus a squared, which is 3a squared, so AD equals a√3.
Now, sin 30° equals BD/AB, which is a/(2a), which is 1/2. Cos 30° equals AD/AB, which is (a√3)/(2a), which is √3/2. Tan 30° equals BD/AD, which is a/(a√3), which is 1/√3. The reciprocals are cosec 30° equals 2, sec 30° equals 2/√3, and cot 30° equals √3.
Similarly, for 60°, sin 60° equals AD/AB, which is (a√3)/(2a), which is √3/2. Cos 60° equals BD/AB, which is a/(2a), which is 1/2. Tan 60° equals AD/BD, which is (a√3)/a, which is √3. The reciprocals are cosec 60° equals 2/√3, sec 60° equals 2, and cot 60° equals 1/√3.
Now, let's understand what happens at 0° and 90°. Consider a right triangle where angle A becomes smaller and smaller. As angle A approaches 0°, the side BC becomes very small, almost zero. The hypotenuse AC becomes almost the same as AB. So, sin A, which is BC/AC, approaches 0, and cos A, which is AB/AC, approaches 1. So we define sin 0° equals 0 and cos 0° equals 1. Then tan 0° equals sin 0°/cos 0°, which is 0/1, which is 0. But cot 0° is not defined because it would be 1/0. Similarly, sec 0° equals 1/cos 0° equals 1, but cosec 0° is not defined because it would be 1/0.
Now, consider when angle A becomes larger and larger, approaching 90°. As angle A approaches 90°, angle C approaches 0°, and side AB becomes very small, almost zero. The hypotenuse AC becomes almost the same as BC. So, sin A approaches 1 and cos A approaches 0. So we define sin 90° equals 1 and cos 90° equals 0. Then tan 90° is not defined, sec 90° is not defined, but cot 90° equals cos 90°/sin 90° equals 0/1 equals 0, and cosec 90° equals 1/sin 90° equals 1.
Let me put all these values in a table for you to remember easily.
For angle 0°, sin is 0, cos is 1, tan is 0, cosec is not defined, sec is 1, and cot is not defined.
For angle 30°, sin is 1/2, cos is √3/2, tan is 1/√3, cosec is 2, sec is 2/√3, and cot is √3.
For angle 45°, sin is 1/√2, cos is 1/√2, tan is 1, cosec is √2, sec is √2, and cot is 1.
For angle 60°, sin is √3/2, cos is 1/2, tan is √3, cosec is 2/√3, sec is 2, and cot is 1/√3.
For angle 90°, sin is 1, cos is 0, tan is not defined, cosec is 1, sec is not defined, and cot is 0.
Now, students, notice an important pattern - as the angle increases from 0° to 90°, sin increases from 0 to 1, and cos decreases from 1 to 0. Tan increases from 0 to infinity as cos approaches 0.
Now, let's look at some examples to see how we can use these values. In Example 6, we have a right triangle ABC, right-angled at B, where AB equals 5 cm and angle ACB equals 30°. We need to find the lengths of BC and AC. Since we know the side opposite to angle C is AB, and the side adjacent to angle C is BC, we can use tan C equals AB/BC. So tan 30° equals 5/BC, which means 1/√3 equals 5/BC, so BC equals 5√3 cm. Now, to find AC, we can use sin C equals AB/AC, so sin 30° equals 5/AC, which means 1/2 equals 5/AC, so AC equals 10 cm. Alternatively, we could have used Pythagoras theorem.
Now, let's move on to trigonometric identities. Students, you may remember that an equation is called an identity when it is true for all values of the variables. Similarly, an equation involving trigonometric ratios is called a trigonometric identity if it is true for all values of the angle involved.
Now, let's prove some important trigonometric identities. Consider a right triangle ABC, right-angled at B. By Pythagoras theorem, we have AB squared plus BC squared equals AC squared. Now, let's divide this entire equation by AC squared. We get (AB/AC)² + (BC/AC)² = (AC/AC)², which is (cos A)² + (sin A)² = 1. So we get cos²A + sin²A = 1. This is true for all angles from 0° to 90°. This is our first trigonometric identity.
Now, let's divide the original equation by AB squared. We get (AB/AB)² + (BC/AB)² = (AC/AB)², which is 1 + tan²A = sec²A. This is our second identity, and it is true for all angles from 0° to less than 90°. At 90°, tan and sec are not defined.
Now, let's divide by BC squared. We get (AB/BC)² + (BC/BC)² = (AC/BC)², which is cot²A + 1 = cosec²A. This is our third identity, and it is true for all angles from greater than 0° to 90°. At 0°, cosec and cot are not defined.
These three identities are very important and you should remember them. They are: sin²A + cos²A = 1 1 + tan²A = sec²A 1 + cot²A = cosec²A
Now, let's see how we can use these identities. Suppose we know that tan A equals 1/√3. Then we can find all other ratios. First, cot A equals √3. Now, using sec²A = 1 + tan²A, we get sec²A = 1 + 1/3 = 4/3, so sec A equals 2/√3. Then cos A equals 1/sec A, which is √3/2. Then sin A equals √(1 - cos²A), which is √(1 - 3/4), which is √(1/4), which is 1/2. Then cosec A equals 2. See how we can find all ratios from just one?
Now, let's look at some examples of proving identities. In Example 10, we need to prove that sec A (1 - sin A)(sec A + tan A) equals 1. Let's start with the left-hand side. Sec A is 1/cos A, and tan A is sin A/cos A. So left-hand side becomes (1/cos A)(1 - sin A)(1/cos A + sin A/cos A), which is (1 - sin A)(1 + sin A)/cos²A, which is (1 - sin²A)/cos²A, which is cos²A/cos²A, which is 1. And that equals the right-hand side. So the identity is proved.
In Example 11, we need to prove that (cot A - cos A)/(cot A + cos A) equals (cosec A - 1)/(cosec A + 1). Let's work on the left-hand side. Cot A is cos A/sin A. So the numerator becomes (cos A/sin A) - cos A, which is cos A(1/sin A - 1). The denominator becomes (cos A/sin A) + cos A, which is cos A(1/sin A + 1). So the left-hand side becomes (1/sin A - 1)/(1/sin A + 1), which is (cosec A - 1)/(cosec A + 1), which is the right-hand side. So the identity is proved.
Now, students, I want to summarize everything we have learned in this chapter.
First, we learned about trigonometric ratios. In a right triangle, for an acute angle A, we defined: sin A = opposite/hypotenuse cos A = adjacent/hypotenuse tan A = opposite/adjacent cosec A = 1/sin A sec A = 1/cos A cot A = 1/tan A
We also learned that tan A = sin A/cos A and cot A = cos A/sin A.
We learned that if we know one trigonometric ratio, we can find all the others using Pythagoras theorem.
We learned the values of trigonometric ratios for specific angles: 0°, 30°, 45°, 60°, and 90°. These are very important and you should memorize them.
We learned that sin and cos are always between 0 and 1, while sec and cosec are always greater than or equal to 1.
Finally, we learned three important trigonometric identities: sin²A + cos²A = 1 1 + tan²A = sec²A 1 + cot²A = cosec²A
These identities are very useful for simplifying expressions and proving other identities.
Students, trigonometry is a very powerful tool in mathematics, and you will be using it in many other chapters and in higher classes as well. So make sure you understand these concepts well and practice as many problems as you can. Thank you for listening so patiently, and I'll see you in the next class. Keep practicing and keep smiling!