Trigonometric Ratios of Quadrantal Angles (`0^\circ`, `90^\circ`, `180^\circ`, `270^\circ`)
For `\theta = 0^\circ`
A point (x, y) = (1, 0) lies on the terminal side of angle `\theta = 0^\circ` as shown below:
In this case, x = 1 & y = 0.
=> Adjacent = 1 & Opposite = 0
According to pythagorean theorem
`text(Adjacent)^2` + `text(Opposite)^2` = `text(Hypotenuse)^2`
=> `text(Hypotenuse)` = `\sqrt(text(Adjacent)^2 + text(Opposite)^2)`
=> Hypotenuse = `\sqrt((1)^2+ (0)^2 )` = `\sqrt(1+0 )` = `\sqrt(1 )` = 1
cos `0^\circ` = `text(Adjacent)/text(Hypotenuse)` = `1/( 1 )` = 1 ; sin `0^\circ` = `text(Opposite)/text( Hypotenuse)` = `0/( 1 )` = 0
tan `0^\circ` = ` text(Opposite)/text( Adjacent )` = `0/( 1 )` = 0 ; cosec `0^\circ` = `text( Hypotenuse )/text( Opposite)` = ` 1/( 0 )` = `\infty`
sec `0^\circ` = `text( Hypotenuse )/text( Adjacent)` = ` 1/( 1 )` = 1 ; cot `0^\circ` = `text(Adjacent)/text( Opposite )` = `1/( 0 )` = `\infty`
For `\theta` = `90^\circ`
A point (x, y) = (0, 1) lies on the terminal side of angle `\theta` = `90^\circ` as shown below:
In this case, x = 0 & y = 1
=> Adjacent = 0 & Opposite = 1
According to pythagorean theorem
`text(Adjacent)^2` + `text(Opposite)^2` = `text(Hypotenuse)^2`
=> `text(Hypotenuse)` = `\sqrt(text(Adjacent)^2 + text(Opposite)^2)`
=> Hypotenuse = `\sqrt((0)^2+ (1)^2 )` = `\sqrt(0+1 )` =` \sqrt(1 ) ` = 1
cos `90^\circ` = ` (Adjacent)/text( Hypotenuse)` = `0/( 1 )` = 0 ; sin `90^\circ` = `text(Opposite)/text( Hypotenuse)` = `1/( 1 )` = 1
tan `90^\circ` = `(Opposite)/( Adjacent )` = `1/( 0 )` = `\infty` ; cosec `90^\circ` = `text( Hypotenuse )/( Opposite)` = `1/( 1 )` = 1
sec `90^\circ` = `text( Hypotenuse )/( Adjacent)` = `1/( 0 )` = `\infty` ; cot `90^\circ` = `text(Adjacent)/( Opposite )` = ` 0/( 1 )` = 0
For `\theta` = `180^\circ`
A point (x, y) = (-1, 0) lies on the terminal side of angle `\theta` = `180^\circ` as shown below:
In this case, x = -1 & y = 0
=> Adjacent = -1 & Opposite = 0
According to pythagorean theorem
`(Adjacent)^2` + `(Opposite)^2` = `text(Hypotenuse)^2`
=> `text(Hypotenuse)` = `\sqrt(text(Adjacent)^2 + text(Opposite)^2)`
=> Hypotenuse = `\sqrt((-1)^2+ (0)^2 ) ` = `\sqrt(1+0 )` = `\sqrt(1 )` = 1
cos `180^\circ` = `text(Adjacent)/text( Hypotenuse)` = `(-1)/( 1 )` = -1 ; sin180° = `text(Opposite)/text( Hypotenuse)` = `0/( 1 )` = 0
tan `180^\circ` = `text(Opposite)/text( Adjacent )` = `0/(-1 )` = 0 ; cosec `180^\circ` = `text( Hypotenuse )/text( Opposite)` = `1/( 0 )` = `\infty`
sec `180^\circ` = `text( Hypotenuse )/text( Adjacent)` = `1/(-1 )` = -1 ; cot `180^\circ` = ` text(Adjacent)/( Opposite )` = `(-1)/( 0 )` = `\infty`
For `\theta` = `270^\circ`
A point (x, y) = (0, -1) lies on the terminal side of angle `\theta` = `270^\circ` as shown below:
In this case, x = 0 & y = -1
=> Adjacent = 0 & Opposite = -1
According to pythagorean theorem
`(Adjacent)^2` + `(Opposite)^2` = `text(Hypotenuse)2`
=> `text(Hypotenuse)` = `\sqrt(text(Adjacent)^2 + text(Opposite)^2)`
=> Hypotenuse = `\sqrt((0)^2+ (-1)^2 )` = `\sqrt(0+1 )` = `\sqrt(1 ) ` = 1
cos `270^\circ` = ` text(Adjacent)/text( Hypotenuse)` = `0/( 1 )` = 0 ; sin `270^\circ` = `text(Opposite)/text( Hypotenuse)` = `(-1)/( 1 )` = -1
tan `270^\circ` = `text(Opposite)/( Adjacent )` = ` (-1)/( 0 )` = `\infty` ; cosec `270^\circ` = `text( Hypotenuse )/( Opposite)` = `1/(-1 )` = -1
sec `270^\circ` = `text( Hypotenuse )/( Adjacent)` = `1/( 0 )` = `\infty` ; cot `270^\circ` = ` text(Adjacent)/( Opposite ) ` = `( 0)/(-1 )` = 0
Trigonometric Ratios of `0^\circ`, `90^\circ`, `180^\circ`, `270^\circ`
The table given below summarizes the trigonometric ratios of angles `0^\circ`, `90^\circ`, `180^\circ`, `270^\circ`
| `\theta` | Cos `\theta` | Sin `\theta` | Tan `\theta` | Cosec `\theta` | Sec `\theta` | Cot `\theta` |
| `0^\circ` | 1 | 0 | 0 | `\infty` | 1 | `\infty` |
| `90^\circ` | 0 | 1 | `\infty` | 1 | `\infty` | 0 |
| `180^\circ` | -1 | 0 | 0 | `\infty` | -1 | `\infty` |
| `270^\circ` | 0 | -1 | `\infty` | -1 | `\infty` | 0 |
Example 1:
Find all trigonometric ratios of 3`\pi`
Solution: We have 3`\pi` = `\pi`+1(2`\pi`)
=> 3`\pi` is coterminal with `\pi` (`180^\circ`). Therefore the trigonometric ratios of 3`\pi` and `\pi` are same.
cos 3`\pi` = cos `\pi` = `text(Adjacent)/text( Hypotenuse)` = `(-1)/( 1 )` = -1
sin 3`\pi` = sin `\pi` = `text(Opposite)/text( Hypotenuse)` = `0/( 1 )` = 0
tan 3`\pi` = tan `\pi` `text(Opposite)/text( Adjacent )` = `0/(-1 )` = 0
cosec 3`\pi` = cosec `\pi` = `text( Hypotenuse )/text( Opposite)` = ` 1/( 0 )` = `\infty`
sec 3`\pi` = sec `\pi` = `text( Hypotenuse )/text( Adjacent)` = `1/(-1 )` = -1
cot 3`\pi` = cot `\pi` = `text(Adjacent)/text( Opposite )` = `(-1)/( 0 )` = `\infty`