# 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

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