The word ‘trigonometry’ is derived from the Greek words ‘tri’(meaning three), ‘gon’ (meaning sides) and ‘metron’ (meaning measure). Trigonometry is the study of relationships between the sides and angles of a triangle.
Applications of trigonometry: There are an enormous number of uses of trigonometry and trigonometric functions.
Early astronomers used it to find out the distances of the stars and planets from the Earth.
In geography to measure distances between landmarks, and in satellite navigation systems.
Even today, most of the technologically advanced methods used in Engineering and Physical Sciences are based on trigonometrical concepts.
The sine and cosine functions are fundamental to the theory of periodic functions such as those that describe sound and light waves.
Basic Trigonometry Rules:
These formulas ONLY work in a right triangle.
The hypotenuse is always across from the right angle.
Questions usually ask for an answer to the nearest units.
You will need a scientific or graphing calculator.
Using Trigonometric Functions to Find a Missing Side
How to set up and solve a trigonometry problem:
Set up the diagram:
Draw a diagram depicting the situation, if one is not given.
Place the angle degrees INSIDE the triangle.
Imagine standing at the reference angle.
LABEL the triangle with o, h, and a. o – opposite side (the side across from you) h – hypotenuse (across from the right angle) a – adjacent side (the leftover side)
“Pair Up” the values. The h pairs with the 25. The o pairs with the x. The a stands alone, which means the a is not involved in the solution to this problem. Cross it out! This problem deals with o and h.
Set up the formula:
Since this problem deals with o and h, we use the sine function since it also deals with o and h.
Replace A with the angle degrees.
Replace o and h with their companion terms.
Use your scientific/graphing calculator to determine the value of sin 42º. (On most graphing calculators, set the mode to degree and press the sin key followed by 42. Most scientific calculators will reverse this order: press 42 first, followed by the sin key.
Solve the equation algebraically. In this problem, cross multiply and solve for x. (When x is on top, you multiply to get the answer. When x is on the bottom, you will divide to get the answer – see example below.)
Round answer to desired value.
Example 2: IIn right triangle ABC, ∠C is the right angle, BC = 17 and angle B = 35º. Find BA to the nearest tenth. Solution: Set up the diagram and the formula in the same manner as was done in Example 1. You should arrive at the drawing and the formula shown here.
Hint: If you are having a problem solving the equation algebraically, remember that when x is on the bottom, you must divide to arrive at your answer. The division is always “divide BY the trig value decimal”.
Hint: Be sure your answer MAKES SENSE!!! The hypotenuse is always the largest side in a right triangle. So, our answer of 26.50 makes sense – it is bigger than the leg of 20.
We have certain trigonometric identities. Like sin2 θ + cos2 θ = 1 and 1 + tan2 θ = sec2 θ etc. Such identities are identities in the sense that they hold for all value of the angles which satisfy the given condition among them and they are called conditional identities.
Example 4: Prove the following identities: (i) cos44 A – cos2 A = sin4 A – sin2 A (ii) cot4 A – 1 = cosec4 A – 2cosec2 A (iii) sin6 A + cos6 A = 1 – 3sin2 A cos2 A. Sol.(i) We have, LHS = cos44 A – cos2 A = cos2A (cos2A – 1) = – cos2 A (1 – cos2 A) = – cos2A sin2A = –(1 – sin2 A) sin2 A = – sin2 A + sin4 A = sin4 A – sin2 A = RHS (ii) We have, LHS = cot4 A – 1 = (cosec2A – 1)2 – 1 [∵ cot2A = cosec2A –1 ⇒ cot4A = (cosec2A – 1)2] = cosec4A – 2 cosec2A + 1 – 1 = cosec4 A – 2cosec2 A = RHS (iii) We have, LHS = sin6 A + cos6 A = (sin2 A)3 + (cos2 A)3 = (sin2 A + cos2 A) {(sin2 A)2 + (cos2 A)2 – sin2 A cos2 A)} [∵ a3 + b3 = (a + b) (a2 – ab + b2)] ={(sin2 A)2 + (cos2 A)2 + 2 sin2 A cos2 A – sin2 A cos2 A} = [(sin2 A + cos2 A)2 – 3 sin2 A cos2 A] = 1 – 3sin2 A cos2 A = RHS
Example 18: Express the ratios cos A, tan A and sec A in terms of sin A. Sol. Since cos2A + sin2A = 1, therefore, cos2A = 1 – sin2A, i.e., cos A = \(\pm \sqrt{1-{{\sin }^{2}}A} \) This gives cos A = \(\sqrt{1-{{\sin }^{2}}A} \) (Why ?) Hence, \( \tan A=\frac{\sin A}{\cos A}=\frac{\sin A}{\sqrt{1-{{\sin }^{2}}A}}\text{ and} \) \( \sec A=\frac{1}{\cos A}=\frac{1}{\sqrt{1-{{\sin }^{2}}A}} \)
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Learn Basic Maths, Topics and Formulas for Mathematics
Mathematics is the only subject that we all will use at every single step of life. If you want to become a Math genius, all you need is a thorough understanding of the concepts of maths like basic calculations, tricks & tips to solve every sum. Students should have a grip on the basic concepts to comprehend their mathematical skills. However, we will learn Maths right from our kindergarten to doctorate courses. Moreover, we have often seen both Math lovers and haters at school time. Students hate the subject because of owing a lot of formulas, equations, figures, and complex calculations. At times, teaching and learning Mathematics become tedious but here comes the best solution to it.
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Frequently Asked Questions
1. Who is the father of mathematics?
Archimedes – The Father Of Mathematics. The father of Archimedes is Astronomer. When he was growing up, the family of Archimedes encouraged him to get an education. Archimedes was passionate in mathematics, science, poetry, politics, and military tactics.
2. What is basic maths?
Basic math is nothing but a simple or basic concept related to mathematics. Usually, counting, addition, subtraction, multiplication, and division are known as basic math operations. The other mathematical concepts are developed on top of the above 4 operations.
3. Which is better basic maths or standard maths?
Standard Maths is nothing but the concepts which students will learn in higher classes, whereas the concepts which are fundamental like adding, subtraction, etc. considered as Basic Mathematics. Therefore, the level of difficulty of the Standard Maths is higher than the Basic Maths. At last, in my view, the conclusion that I can give is “Basic Maths is better than Standard Maths”.
4. What are the Basic Concepts of Mathematics?
The major concepts involved in pure maths are as follows:
Number system,
Pre-algebra,
Algebra,
Commercial Arithmetic,
Geometry,
Trigonometry,
Arithmetic,
Statistics,
Calculus,
Probability and many more
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Anyone can become really good at math. All you need is practice and understanding power while solving math problems. If you follow these 8 tips then no one can stop you to succeed in solving math sums. So, check out the following points and focus on your weak points more while practicing:
Do all of the homework. Don’t ever think of homework as a choice.
Fight not to miss class.
Find a friend to be your study partner.
Establish a good relationship with the teacher.
Analyze and understand every mistake.
Get help fast.
Don’t swallow your questions.
Basic skills are essential.
Algebra I must be mastered.
Understand what the calculator is doing.
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APLUSTOPPER offers students reliable math resources as well as theoretical explanations. A few primary resources that every student may require while learning Maths are listed here and available at APlus Topper for free:
CBSE NCERT Solutions for Maths
RS Aggarwal and RD Sharma Solutions for Class 1 to Class 12 Mathematics
CBSE Maths Previous Question papers for all Classes
Example 5: Express each of the following in terms of trigonometric ratios of angles between 0º and 45º; (i) cosec 69º + cot 69º (ii) sin 81º + tan 81º (iii) sin 72º + cot 72º Sol. (i) We have, cosec 69º + cot 69º = cosec (90º – 21º) + cot (90º – 21º) = sec 21º + tan 21º [∵ cosec (90º – θ) = sec θ and cot (90º –θ) = tan θ] (ii) We have, sin 81º + tan 81º = sin (90º – 9º) + tan (90º – 9º) = cos 9º + cot 9º [∵ sin (90º – θ) = cos θ and tan (90º –θ) = cot θ] (iii) We have, sin 72º + cot 72º = sin (90º – 18º) + cot (90º – 18º) = cos 18º + tan 18º [∵ sin (90º – 18º) = cos 18º and tan (90º –18º) = cot 18º]
Example :7 If tan 2θ = cot (θ + 6º), where 2θ and θ + 6º are acute angles, find the value of θ. Sol. We have, tan 2θ = cot (θ + 6º) ⇒ cot(90º – 2θ) = cot (θ + 6º) ⇒ 90º – 2θ = θ + 6º ⇒ 3θ = 84º ⇒ θ = 28º
Example :8 If A, B, C are the interior angles of a triangle ABC, prove that \( \tan \frac{B+C}{2}=\cot \frac{A}{2} \) Sol. In ∆ABC, we have A + B + C = 180º ⇒ B + C = 180º – A \( \Rightarrow \frac{B+C}{2}=\text{ }90{}^\text{o}-\frac{A}{2} \) \( \Rightarrow \tan \left( \frac{B+C}{2} \right)=\tan \left( 90{}^\text{o}-\frac{A}{2} \right) \) \( \Rightarrow \tan \left( \frac{B+C}{2} \right)=\cot \frac{A}{2} \)
Example :9 If tan 2A = cot (A – 18°), where 2A is an acute angle, find the value of A. Sol. tan 2A = cot (A – 18°) cot (90° – 2A) = cot (A – 18°) (∵ cot (90° – θ) = tan θ) 90° – 2A = A – 18° 3A = 108° A = 36°
Example :10 If tan A = cot B, prove that A + B = 90°. Sol. ∵ tan A = cot B tan A = tan (90° – B) A = 90° – B A + B = 90°. Proved
Example :11 If A, B and C are interior angles of a triangle ABC, then show that \( \sin \left( \frac{B+C}{2} \right)=\cos \frac{A}{2} \) Sol. ∵A + B + C = 180° (a.s.p. of ∆) B + C = 180° – A \(\left( \frac{B+C}{2} \right)=90{}^\circ -\frac{A}{2}\) \( \sin \left( \frac{B+C}{2} \right)=\sin \left( 90{}^\circ -\frac{A}{2} \right) \) \( \sin \left( \frac{B+C}{2} \right)=\cos \frac{A}{2} \) Proved.
Example :12 Express sin 67° + cos 75° in terms of trigonometric ratios of angles between 0° and 45°. Sol. ∵ 23 = 90 – 67 & 15 = 90 – 75 ∴ sin 67° + cos 75° = sin (90 – 23)° + cos (90 – 15)° = cos 23° + sin 15°.