Trigonometric ratios are Sine, Cosine, Tangent, Cotangent, Secant and Cosecant. The standard angles for these trigonometric ratios are 0°, 30°, 45°, 60° and 90°. These angles can also be represented in the form of radians such as 0, Ï€/6, Ï€/4, Ï€/3, and Ï€/2. These angles are most commonly and frequently used in trigonometry. Learning the values of these trigonometry angles is very necessary to solve various problems.
Trigonometric Ratios Formulas:
The six trigonometric ratios are basically expressed in terms of the right-angled triangle.
∆ABC is a right-angled triangle, right-angled at B (shown in figure 1). The six trigonometric ratios for ∠C are defined as:
sin ∠C = ABAC
cosec ∠C = 1sin ∠C
cos ∠C = BCAC
sec ∠C = 1cos ∠C
tan ∠C = sin ∠Ccos ∠C
cot ∠C = 1tan ∠C
The standard angles for which trigonometric ratios can be easily determined are 0°,30°,45°,60° and 90°. The values are determined using properties of triangles. The two acute angles of a right-angled triangle are complementary.
Trigonometric Ratios Table (Standard Angles)
Angle = ∠C | 0° | 30° | 45° | 60° | 90° |
sin C | 0 | 12 | 12√ | 3√2 | 1 |
cos C | 1 | 3√2 | 12√ | 12 | 0 |
tan C | 0 | 13√ | 1 | 3–√ | Not Defined |
cosec C | Not Defined | 2 | 2–√ | 23√ | 1 |
sec C | 1 | 23√ | 2–√ | 2 | Not Defined |
cot C | Not Defined | 3–√ | 1 | 13√ | 0 |
The above table shows the important angles for all the six trigonometric ratios. Let us learn here how to derive these values.
Derivation of Trigonometric Ratios for Standard Angles
Value of Trigonometric Ratios for Angle equal to 45 degrees
In ⊿ABC, if ∠C = 45°, then ∠A = 45°. Since the angles are equal, ⊿ABC becomes a right angled isosceles triangle. So, AB = BC. Assume AB = BC = a units.
Using Pythagoras theorem ,
AC2 = AB2 + BC2
AC2 = a2 + a2
AC = a2–√ units
∠C = 45°
∴ sin ∠C = sin 45° = ABAC = aa2√ = 12√
cosec 45° = 1sin 45° = 2–√
cos ∠C = cos 45° = BCAC = aa2√ = 12√
sec 45° = 1cos 45° = 2–√
tan 45° = sin 45°cos 45° = a2√a2√ = 1
cot 45° = 1tan 45° = 1
Value of Trigonometric Ratios for Angle equal to 30 and 60 degrees
In figure 3, ΔPQR is equilateral. The perpendicular from any vertex on the opposite side is coincident with the angle bisector of that particular vertex. Also, the perpendicular bisects the opposite side. If a perpendicular PS is dropped on QR, then ∠QPS = ∠SPR = 30° and QS = SR. Assume PQ = QR = RP = 2a units.
⇒ QS = SR = a units
In ΔPSQ, by Pythagoras theorem,
PQ2 = QS2 + PS2
PS2 = (2a)2 − a2
PS = 3a2−−−√ = 3–√a
∠SPQ = 30°
sin ∠SPQ = sin 30° = SQPQ = a2a = 12
cosec 30° = 1sin 30° = 2
cos ∠SPQ = cos 30° = PSPQ = 3√a2a = 3√2
sec 30° = 1cos 30° = 23√
tan 30° = sin 30°cos 30° = 123√2 = 13√
cot 30° = BCAB = 3–√
Similarly, ratios of 60° are determined by finding the ratios of ∠SQP as
sin 60° = 3√2
cos 60° = 12
tan 60° = 3–√
cot 60° = 13√
cosec 60° = 23√
sec 60° = 2
Value of Trigonometric Ratios for Angle equal to 0 and 90 degrees
In ΔABC is a right angled triangle. If the length of side BC is continuously decreased, then value of ∠A will keep on decreasing. Similarly, value of ∠C is increasing as length of BC is decreasing. When BC = 0, ∠A = 0 , ∠C = 90° and AB = AC.
Taking ratios for ∠A = 0°
sin ∠A = sin 0° = BCAC = 0AC = 0
cosec 0° = 1sin 0° = 10 = Not Defined
cos ∠A = cos 0° = ABAC = ACAC = 1
sec 0° = 1cos 0° = 11 = 1
tan 0° = sin 0°cos 0° = 01 = 0
cot 0° = 1tan 0° = 10 = Not Defined
Taking ratios for ∠C = 90°
sin ∠C = sin 90° = ABAC = ACAC = 1
cosec90° = 1sin 90° = 11 = 1
cos ∠C = cos 90° = BCAC = 0AC = 0
sec 90° = 1cos 90° = 10 = Not Defined
tan 90° = sin 90°cos 90° = 10 = Not Defined
cot 90° = 1tan 90° = 01 = 0
Following is the trigonometric ratios table which contains all the trigonometric ratios of standard angles:
Solved Examples
Question 1: What is the value of tan 30+sin 60?
Solution: tan 30 = 1/√3 and sin 60 = √3/2
Adding both the values we get;
1/√3 + √3/2
Rationalising the denominator gives:
(2+√3.√3)/2√3
2+3/2√3
5/2√3
Question 2: What is the value of sin45 – cos 45?
Solution: Sin 45 = 1/√2 and cos 45 = 1/√2
Therefore, on putting the values we get:
1/√2 – 1/√2 = 0