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Trigonometry values are the values of standard angles for a given right-angled triangle with respect to trigonometric ratios. The value of different trigonometric ratios such as Sine, Cosine, Tangent, Secant, Cosecant and Cotangent for standard angles (0°, 30°, 45°, 60° and 90°) are commonly used to solve trigonometric problems. The trigonometry values have various applications in the fields like architecture, engineering, oceanography etc. Here, you will learn about trigonometric values, trigonometric ratios and some important formulas along with solved examples.
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Trigonometry Values Table
Let ABC is a right-angled triangle with right-angled at C. This triangle has a hypotenuse AB, an adjacent side AC which is adjacent to ∠CAB and a perpendicular BC opposite to ∠CAB.
Here are the trigonometry values for angles 0°, 30°, 45°, 60° and 90° with respect to trigonometric functions Sin, Cos, Tan, Sec, Cosec and Cot.
| Angle | 0° | 30° | 45° | 60° | 90° |
| Sin∏ | 0 | 1/2 | 1/√2 | √3/2 | 1 |
| Cos∏ | 1 | √3/2 | 1/√2 | 1/2 | 0 |
| Tan∏ | 0 | 1/√3 | 1 | √3 | ∞ |
| Cosec∏ | ∞ | 2 | √2 | 2/√3 | 1 |
| Sec∏ | 1 | 2/√3 | √2 | 2 | ∞ |
| Cot∏ | ∞ | √3 | 1 | 1/√3 | 0 |
Some very important observations from the table are
The value of theta increases from 0° to 90° for Sin∏
The value of theta decreases from 0° to 90° for Cos∏
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Trigonometric Ratios (T-Ratios)
Let us take a right-angled triangle ABC right angled at B.
As we can see, AB is the perpendicular, BC is the base and AC is the hypotenuse in the given triangle ABC.
The trigonometric ratios for angle B in the triangle ABC are given below
| Functions | Abbreviation | Relationship to sides in the given triangle ABC | Ratios |
|---|---|---|---|
| Sine Function | Sin B | Perpendicular/Hypotenuse | AB/AC |
| Cosine Function | Cos B | Base/ Hypotenuse | BC/AC |
| Tangent Function | Tan B | Perpendicular/Base | AB/BC |
| Cosecant Function | Cosec B | Hypotenuse/Perpendicular | AC/AB |
| Secant Function | Sec B | Hypotenuse/Base | AC/BC |
| Cotangent Function | Cot B | Base/Perpendicular | BC/AB |
An easy way to learn the above trigonometric ratios is
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Sin ∏ = Perpendicular/Hypotenuse
Cos ∏ = Base/Hypotenuse
Tan ∏ = Perpendicular/Base
Also, the other three trigonometric functions that are Cosec, Sec and Cot have reciprocal relationships with Sin, Cos and Tan, respectively. So,
Cosec∏ = 1/Sin∏
Sec∏ = 1/Cos∏
Cot∏ = 1/Tan∏
Note: Tan∏ = Sin∏/Cos∏ and Cot∏ = Cos∏/Sin∏
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Trigonometry Ratios Formula
- Tan ∏ = sin ∏ /cos ∏
- Cot ∏ = cos ∏ /sin ∏
- Sin ∏ = 1/cosec ∏
- Cos ∏ = sin ∏ /tan ∏ = 1/sec ∏
- Sec ∏ = tan∏ /sin ∏ = 1/cos ∏
- Cosec ∏ = 1/sin ∏
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Trigonometric Identities
The three basic identities of trigonometry are as follows
- Cos2∏ + Sin2∏ = 1
- Tan2∏ + 1 = Sec2∏
- Cot2∏ + 1 = Cosec2∏
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Trigonometric Ratios of Complementary Angles
Complementary angles are those angles whose sum is 90°.
Trigonometric Ratios of Complementary Angles are
- Sin (90° - ∏) = Cos∏
- Sec (90° - ∏) = Cosec∏
- Cos (90° - ∏) = Sin∏
- Cosec (90° - ∏) = Sec∏
- Tan (90° - ∏) = Cot∏
- Cot (90o - ∏) = Tan∏
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Things to Remember Based on Trigonometry Values
- Trigonometry values are the values of standard angles for a given right-angled triangle with respect to trigonometric ratios.
- The value of theta increases from 0° to 90° for Sin∏.
- The value of theta decreases from 0° to 90° for Cos∏.
- Trigonometric Ratios-
- Sin∏ = Perpendicular/Hypotenuse
- Cos ∏ = Base/Hypotenuse
- Tan ∏ = Perpendicular/Base
- Cosec∏ = 1/Sin∏ = Hypotenuse/ Perpendicular
- Sec∏ = 1/Cos∏ = Hypotenuse/ Base
- Cot∏ = 1/Tan∏ = Base/ Perpendicular
- Some basic trigonometric identities-
- Cos2∏ + Sin2∏ = 1
- Tan2∏ + 1 = Sec2∏
- Cot2∏ + 1 = Cosec2∏
- The value of Tan∏ is always less than 1.
- The value of Sin∏ or Cos∏ never exceeds 1.
- The value of Sec∏ or Cosec∏ is always greater than 1 or equal to 1.
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Important Questions Based on Trigonometry Values
Ques: Evaluate Sin 25o Cos 65o + Cos 25o Sin 65o (2 Marks)
Ans: Sin ( 90 – 65 )o Cos 65o + Cos ( 90 – 65 )o Sin 650
=Cos 65o Cos 65o + Sin 65o Sin 65o
Cos2 65o + Sin265o = 1
Ques: Evaluate 1 – Tan245 / 1 + Tan2 45 (2 Marks)
Ans: Since,
Tan 45o = 1
Therefore, 1-1/1+1
= 0/2
= 0
Ques: If ∠A and ∠B are acute angles such that Cos A = Cos B, then show that ∠A= ∠B. (2 Marks)
Ans: Since Cos A = Cos B (given)
Therefore, in Triangle ABC
AC/AB = BC/AB
∠A= ∠B (Angles opposite to equal sides are equal)
Ques: Evaluate Sin 60° Cos 30° + Sin 30° Cos 60° (2 Marks)
Ans: We know that:
Sin 60°= √3/ 2
Cos 30°= √3/2
Sin 30°= 1/2
Cos 60°= 1/2
So,
Sin 60° Cos 30° + Sin 30° Cos 60° = √3/ 2 . √3/2 + 1/2.1/2
4/4 = 1
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Ques: Express Sin 67° + Cos 75° in terms of trigonometric ratios of angles between 0° and 45° (2 Marks)
Ans: Sin 67° + Cos 75°
= Sin (90 – 23)° + Cos (90 – 27)°
= Cos 23° +Sin 25°
So, Sin 67° + Cos 75° can also be expressed as Cos 23° + Sin 25°
Ques: If Sec 4A = Cosec ( A – 20° ) , where 2A is an acute angle, find the value of A. (2 Marks)
Ans: Sec 4A = Cosec ( A - 20° )
Cosec ( 90 - 4A ) = Cosec (A- 20° )
90 – 4A = A – 20
110 = 5A
A = 22°
Ques: Prove that: 9 sec2A – 9 tan2A = 9 (2 Marks)
Ans: Taking 9 common,
We get, 9 ( sec2A – tan2A )
Using identity, sec2 A = 1 + tan2 A
9 ( 1 + tan2A – tan2 A ) = 9
Hence, proved.
Ques: Prove that ( Sec A + Tan A ) ( 1 – Sin A ) = Cos A (3 Marks)
Ans: ( 1/Cos A + Sin A / Cos A ) ( 1 – Sin A )
1 – Sin2A / Cos A
Cos2 A / Cos A = Cos A
Hence Proved
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Ques: Prove that: ( Sin A + Cosec A )2 + ( Cos A + Sec A )2 = 7 + Tan2A + Cot2A (3 Marks)
Ans: L.H.S
Sin2A + \(\frac{1}{Sin^2 A}\) + 2 + Cos2A + \(\frac{1}{Cos^2 A}\) + 2
=5 + Cosec2A + Sec2A
=5 + (1 + Cot2A) + (1 + Tan2A)
=7 + Cot2A + Tan2A
= R.H.S
Hence, proved.
Ques: If A, B and C are interior angles of a triangle ABC, then show that Sin ( B+C/2 ) = Cos A/2 (3 Marks)
Ans: To Prove: Sin ( B+C/2 ) = Cos A/2
Proof: A + B + C = 180o (Angle Sum Property)
B + C = 180o – A
Divide the equation by 2 both sides
B+C/2 = 90o – A/2
Applying Sin both sides
Sin (B+C/2) = Sin (90o – A/2)
Therefore, Sin ( B+C/2 ) = Cos A/2 (Sin (90o - ∏) = Cos ∏ )
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Ques: Express the trigonometric ratios Sin A, Sec A and Tan A in terms of Cot A. (5 Marks)
Ans: (I) Sin A = 1/ \(\sqrt{1+Cot^2A}\)
(II) Sec A = \(\sqrt{1+Tan^2A}\)( Using identity Sec2A = 1 + Tan2A )
Therefore, Sec A = \(\sqrt{1+ \frac{1}{Cot^2A}}\) ( Tan A = 1/Cot A )
Taking L.C.M
Hence, Sec A = \(\sqrt{\frac{1+Cot^2A}{Cot A}}\)
(III) Tan A = 1/Cot A
Ques: Prove that \(\frac{Cot A-Cos A}{Cot A+Cos A}\) = \(\frac{Cosec A-1}{Cosec A+1}\) (5 Marks)
Ans: L.H.S
\(\frac{\frac{Cos A}{Sin A} - Cos A}{\frac{Cos A}{Sin A} + Cos A}\)
Taking L.C.M in the numerator and denominator separately, we get
\(\frac{\frac{Cos A - Cos A. Sin A}{Sin A}}{\frac{Cos A + CosA.SinA}{Sin A}}\)
Therefore, \(\frac{Cos A ( 1-Sin A )}{Cos A ( 1+Sin A )}\)
Hence, \(\frac{1-Sin A}{1+Cos A}\)
RHS
\(\frac{\frac{1}{Sin A} - 1}{\frac{1}{Sin A} + 1}\)
On taking L.C.M, We get
\(\frac{1-Sin A}{1+Sin A}\)
Therefore, L.H.S = R.H.S
Hence, Proved
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