Measurement of Angles
There are three systems for measuring the angles, which are given below
(i) Sexagesimal System (Degree Measure)
In this system, a right angle is divided into 90 equal parts, called the degrees. The symbol 1º is used to denote one degree. Each degree is divided into 60 equal parts, called the minute and one minute is divided into 60 equal parts, called the seconds. Symbols 1′ and 1” are used to denote one minute and one second respectively.
i.e. 1 right angle = 90º,
1º = 60′ and 1′ = 60”
(ii) Circular System (French System)
In this system, the angle is measured in radians. A radian is an angle subtended at the centre of a circle by an arc, whose length is equal to the radius of the circle. The number of radians in an angle subtended by an arc of a circle at the centre is equal to arc\over radius
(iii) Centesimal System (French System)
In this system, a right angle is divided into 100 equal parts, called the grades. Each grade is subdivided into 100 min and each minute is divided into 100 s.
i.e. 1 right angle = 100 grades = 100g
1g = 100′, 1′ = 100”
Relation between Degree and Radian
- π radian = 180º
- 1 radian = 180º\over π
- 1º = π\over 180º
Length of an Arc of a Circle
If in a circle of radius r, an arc of length l subtend an angle θ radian at the centre, then
length of the arc (l) = θ×r
Trigonometric Ratio
(i) sin θ = \text{Perpendicular}\over \text{Hypotenuse}
(ii) cos θ = \text{Base}\over \text{Hypotenuse}
(iii) tan θ = \text{Perpendicular}\over \text{Base}
(iv) cot θ = \text{Base}\over \text{Perpendicular}
(v) sec θ = \text{Hypotenuse}\over \text{Base}
(vi) cosec θ = \text{Hypotenuse}\over \text{Perpendicular}
Quotient Relation
(i) tan θ = \text{sin\ θ}\over \text{cos\ θ}
(ii) cot θ = \text{cos\ θ}\over \text{sin\ θ}
(iii) sec θ = 1\over \text{cos\ θ}
(iv) cosec θ = 1\over \text{sin\ θ}
Square Relation
| (i) sin2θ+cos2θ=1 | (ii) sec2θ-tan2θ=1 |
| (a) sin2θ=1-cos2θ | (a) sec2θ=1+tan2θ |
| (b) cos2θ=1-sin2θ | (b) tan2θ=sec2θ-1 |
| (iii) cosec2θ-cot2θ=1 | |
| (a) cosec2θ=1+cot2θ | |
| (b) cot2θ=cosec2θ-1 | |
Trigonometric Function with Angles
| Angle | 0º | 30º | 45º | 60º | 90º |
| sin | 0 | 1/2 | 1/√2 | √3/2 | 1 |
| cos | 1 | √3/2 | 1/√2 | 1/2 | 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 |
Compound Angles
(i) sin(A+B)=sinAcosB+cosAsinB
(ii) sin(A-B)=sinAcosB-cosAsinB
(iii) cos(A+B)=cosAcosB−sinAsinB
(iv) cos(A-B)=cosAcosB+sinAsinB
(v) tan(A+B)=\text{tanA+tanB}\over \text{1-tanAtanB}
(vi) tan(A-B)=\text{tanA-tanB}\over \text{1+tanAtanB}
(vii) cot(A+B)=\text{cotAcotB-1}\over \text{cotA + cotB}
(viii) cot(A-B)=\text{cotAcotB}+1\over \text{cotA - cotB}
Some Important Result
(i) sin(A+B)sin(A-B)=sin2A-sin2B=cos2B-cos2A
(ii) cos(A+B)cos(A-B)=cos2A-sin2B=cos2B-sin2A
Product to Sum or Differnce
(i) 2sinAcosB = sin(A+B)+sin(A-B)
(ii) 2cosAsinB = sin(A+B)-sin(A-B)
(iii) 2cosAcosB = cos(A+B)+cos(A-B)
(iv) 2sinAsinB = cos(A-B)-cos(A+B)
Sum or Difference to Product
(i) sinC + sinD = 2sin({\text{C+D}\over\text{2}})cos({\text{C-D}\over\text{2}})
(ii) sinC – sinD = 2cos({\text{C+D}\over\text{2}})sin({\text{C-D}\over\text{2}})
(iii) cosC + cosD = 2cos({\text{C+D}\over\text{2}})cos({\text{C-D}\over\text{2}})
(iv) cosC – cosD = – 2sin({\text{C+D}\over\text{2}})sin({\text{C-D}\over\text{2}})
Multiple Angles
| (i) sin 2θ | = 2sinθcosθ |
| = 2\text{tan θ}\over 1+\text{tan}^2\text{θ} | |
| (ii) cos 2θ | = cos2θ-sin2θ |
| = 2cos2θ-1 | |
| = 1-2sin2θ | |
| = 1-\text{tan}^2\text{θ}\over 1+\text{tan}^2\text{θ} | |
| (iii) tan 2θ | = 2\text{tan θ}\over 1-\text{tan}^2\text{θ} |
| (iv) sin 3θ | = 3 sin θ – 4sin3 θ |
| (v) cos 3θ | = 4cos3 θ – 3cos θ |
| (vi) tan 3θ | = 3\text{tan θ} - \text{tan}^3\text{θ}\over 1-3\text{tan}^2\text{θ} |
Sub-Multiple Angles
| (i) sin θ | = 2sin(θ/2)cos(θ/2) |
| = 2\text{tan (θ/2)}\over 1+\text{tan}^2\text{(θ/2)} | |
| (ii) cos θ | = cos2(θ/2)-sin2(θ/2) |
| = 2cos2(θ/2)-1 | |
| = 1-2sin2(θ/2) | |
| = 1-\text{tan}^2\text{(θ/2)}\over 1+\text{tan}^2\text{(θ/2)} | |
| (iii) tan θ | = 2\text{tan (θ/2)}\over 1-\text{tan}^2\text{(θ/2)} |
| (iv) sin θ | = 3 sin (θ/3) – 4sin3 (θ/3) |
| (v) cos θ | = 4cos3 (θ/3) – 3cos (θ/3) |
| (vi) tan θ | = 3\text{tan (θ/3)} - \text{tan}^3\text{(θ/3)}\over 1-3\text{tan}^2\text{(θ/3)} |
General Solution
| (i) sin θ = 0 | θ = nπ, where n ∈ z |
| (ii) cos θ = 0 | θ = (2n+1)π/2, where n ∈ z |
| (iii) tan θ = 0 | θ = nπ, where n ∈ z |
| (iv) sin θ = sin α | θ = nπ + (-1)nα , where n ∈ z |
| (v) cos θ = cos α | θ = 2nπ ± α, where n ∈ z |
| (vi) tan θ = tan α | θ = nπ + α, where n ∈ z |
| (vii) sin2 θ = sin2 α | θ = nπ ± α, where n ∈ z |
| (viii) cos2 θ = cos2 α | θ = nπ ± α, where n ∈ z |
| (ix) tan2 θ = tan2 α | θ = nπ ± α, where n ∈ z |
| (x) sin θ = 1 | θ = (4n+1)π/2, where n ∈ z |
| (xi) cos θ = 1 | θ = 2nπ, where n ∈ z |
