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:

Quadrantal-Angles

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:

Quadrantal-Angles

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:

Quadrantal-Angles

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:

Quadrantal-Angles

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`






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