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The measurement of the three angles of a triangle is discussed in the branch of mathematics known as trigonometry. Trigonometry is always useful in arithmetic and is also used in various branches of science. We are going to go into more detail about this triangulation. Our famous Indian mathematician "Aryabhata" whom we know as the inventor of zero, was the first to apply trigonometry.
Read Also: Class 10 Introduction to Trigonometry
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Keyterms: Triangle, Trigonometry, zero, Base, right angle, Perpendicular, adjacent
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Basics of Trigonometry
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The part of the trigonometry figure that gives us the relation between the side of the triangle and the angle of the triangle. Trigonometry is founded by three Greek words. Those are- ‘tri’, ‘gon’ and ‘metro’. Here ‘tri’ means three, ‘gon’ means sides and ‘metro’ means to measure. This means the measurement of the three sides. The angle formed by combining these three sides is called a triangle. So, we will discuss the side of a triangle and the angle of the triangle. The technique we use to calculate the height and the distance is called trigonometry. One of these is a right angle triangle whose angle is always at a 90-degree angle. When it comes to trigonometry, keep in mind, a triangle is always imagined as a triangle. The sum of the three angles of any triangle is equal to 180 degrees.
Now we will know the details about the picture above. The right angle in the picture is B. If we make A an angle, then the side in front of the angle is called side opposite to ∠A. We call this Perpendicular. The one on which a triangle is located is called Base. Here Base is AB. There is another definition of this. Base can be called side adjacent to ∠A. That is, the side combined with angle. So, ∠A=BC. The longest side of the triangle is the hypotenuse. To memorize the triangular formula easily, one small mantra has to be memorized, which is-
Check Important Questions for Introduction to Trigonometry
| sin\(\theta\) | cos\(\theta\) | tan\(\theta\) |
| Pandit (P) | Badri (B) | Parsad (P) |
| Hari (H) | Hari (H) | Bole (B) |
| cosec\(\theta\) | sec\(\theta\) | cot\(\theta\) |
Trigonometric Ratios
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| sin\(\theta\) | \(\frac{P}{H}\) | \(\frac{BC}{AC} = \frac{ \text {side opposite to }\angle A } {hypotenuse}\) |
| cos\(\theta\) | \(\frac{B}{H}\) | \(\frac{AB}{AC} = \frac {\text {side adjacent to }\angle A}{hypotenuse}\) |
| tan\(\theta\) | \(\frac{P}{B}\) | \(\frac{BC}{AB} = \frac { \text {side opposite to }\angle A}{\text {side adjacent to }\angle A}\) |
| cosec\(\theta\) | \(\frac{H}{P}\) | \(\frac{AC}{BC} = \frac {hypotenuse}{\text {side opposite to }\angle A}\) |
| sec\(\theta\) | \(\frac{H}{B}\) | \(\frac{AC}{AB} = \frac {hypotenuse}{ \text{side adjacent to }\angle A}\) |
| cot\(\theta\) | \(\frac{B}{P}\) | \(\frac{AB}{BC} = \frac {\text{side adjacent to }\angle A}{\text{side opposite to }\angle A}\) |
Read More Formulas for Trigonometry
So we understand that the value of cosecθ is the opposite of the value of sinθ, similarly the value of secθ is the opposite of cosθ and the value of cotθ is the opposite of tanθ.
Now we will explain through Hexagon:
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Tanθ = \(\frac {sin\theta}{cos\theta}\)
Sinθ = \(\frac {cos\theta}{cot\theta}\)
Cosθ = \(\frac {cot\theta}{cosec\theta}\)
Cotθ = \(\frac {cosec\theta}{sec\theta}\)
Cosecθ = \(\frac {sec\theta}{tan\theta}\)
secθ = \(\frac {tan\theta}{sin\theta}\)
Read Also: Trigonometric Identities
Tanθ = \(\frac {sec\theta}{cosec\theta}\)
Sinθ = \(\frac {tan\theta}{sec\theta}\)
Cosθ = \(\frac {sin\theta}{tan\theta}\)
Cotθ = \(\frac {cos\theta}{sin\theta}\)
Cosecθ = \(\frac {cot\theta}{cos\theta}\)
secθ = \(\frac {cosec\theta}{cot\theta}\)
Next,
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The two that are facing each other here, those two multiplications will be 1. So,
Sinθ * Cosecθ = 1
Tanθ * Cotθ = 1
secθ * Cosθ = 1
Next,
Check Further: Trigonometric Functions
The function between the two functions will be the multiplication of those two functions. So,
Tanθ * Cosθ = Sinθ
Sinθ * Cotθ = Cosθ
Cosθ * Cosecθ = Cotθ
Cotθ * secθ = Cosecθ
Cosecθ * Tanθ = secθ
Secθ * Sinθ = Tanθ
Next,
Read Further: Class 11 Permutations and Combinations
We all know that the Complementary angles are sum of 90-degree. So, the new formulas are,
Sinθ = Cos (90 - θ)
Tanθ = Cot (90 - θ)
Secθ = Cosec (90 - θ)
Cosθ = Sin (90 - θ)
Cotθ = Tan (90 - θ)
Cosecθ = Sec (90 - θ)
Square Formula (clockwise)
Read More: Class 11 Pascal’s Triangle
We are taking three triangles here. So, the new formulas are,
Sin2θ + Cos2θ = 1
1 + Cot2θ = Cosec2θ
Tan2θ + 1 = Sec2θ
Square Formula (anti-clockwise)
1 - Cos2θ = Sin2θ
Cosec2θ - Cot2θ = 1
Sec2θ – 1 = Tan2θ
Also Read:
| Related Articles | ||
|---|---|---|
| Types of Matrices | Invertible Matrices | |
| Identity Matrix | Transpose of a Matrix | |
| Scalar Matrix | Matrices | |
The formula for trigonometry through the table
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| Angle | 0° | 30° | 45° | 60° | 90° |
|---|---|---|---|---|---|
| Sin | 0 | 12 | 1√2 | √32 | 1 |
| Cos | 1 | √32 | 1√2 | 12 | 0 |
| Tan | 0 | 1√3 | 1 | √3 | ∞ |
| Cot | ∞ | √3 | 1 | 1√3 | 0 |
| Sec | 1 | 2√3 | √2 | 2 | ∞ |
| Cosec | ∞ | 2 | √2 | 2√3 | 1 |
Check Important Notes for Inverse Trigonometric Functions
Sample Questions
Ques. Evaluate sin2 60 + 2 tan 45°- cos 2 30° (2 marks)
Ans: \((\frac{\sqrt{3}}{2})^2\) + 2 x 1 – \((\frac{\sqrt{3}}{2})^2\)
= \(\frac{3}{4}\) + 2 – \(\frac{3}{4}\)
= 2
Ques. If tan2A = cot (A - 18°), where 2A is acute angel, fine the value of A. (2 marks)
Ans. =) tan2A = tan 90- A-18°
=) tan2A = tan 108°-A
=) 2A = 108° – A
=) 3A = 108°
=) A = 36°
Ques. Let S = sin230° + sin245° + sin260° and P = cosec245°. sec230°. sin390°. cos60° then the correct statement is. a) S<P, b) S = P, c) SP = 2, d) S + P>3 (2 marks)
Ans.
Correct option is C)S = \(\frac{1}{4}+\frac{1}{2}+\frac{3}{4}\) = \(\frac{3}{2}\)
P = 2 × \(\frac{4}{3}\) × 1 × \(\frac{1}{2}\) = \(\frac{4}{3}\)
⇒ S > P
Now, SP = \(\frac{3}{2}\) × \(\frac{4}{3}\) = 2
S + P = \(\frac{3}{2}\) + \(\frac{4}{3}\)= \(\frac{9 + 8}{6} = \frac{17}{6}\) < 3
Hence, option C is correct.
So, SP = 2
Ques. If sin x + sin2x = 1 then the value of cos2x + cos4x is equal to – a) 0, b) 1, c) 2, d)
3 (2 marks)Ans. Correct option is b)
sin x = 1 - sin2x
sin x = cos2x
So,
cos2x = sin x
cos4x = sin2x
cos2x + cos4x = 1
Ques. The expression 2(1 + cosx) – sin2x is same as: A) (1 – cosx)2, B) 1 - cos2x, C) (1 + cosx)2, D) 1 + cos2x (2 marks)
Ans. = 2(1 + cosx) – sin2x
= 2 + 2cosx – (1 + cos2x)
= 1 + 2cosx + cos2x
= (1 + cosx)2
Ques. Prove that: (sinθ + cosθ) (tanθ + cotθ) = secθ + cosecθ (2 marks)
Ans:
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