Trigonometric Ratios of Standard Angles in Third and Fourth Quadrants

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Trigonometric Ratios of 210o or (7π/6)c:

Let m∠ AOP = θ = 210o = (7π/6)c

m∠ POM = 30o

Ray OA is the initial arm of the angle.



The terminal arm of the angle ray OP intersects the circle at P(x, y)

Let PM be perpendicular to OX’. Thus ΔOMP is 30o-60o-90o triangle

PM = 1/2(OP) = 1/2 (1) = 1/2  (side opposite to 30o)

OM = 3/2(OP) = 3/2 (1) = 3/2  (side opposite to 60o)



Point P is in the third quadrant

Hence x = –3/2 an y = -1/2. Thus

sin 210o = y = – 1/2

cos 210o =  x = – 3/2

tan 210o =  y/x  = (-1/2)/(-3/2) = 1/3



cosec 210o = 1/y = 1/(-1/2) = – 2

sec 210o  =  1/x = 1/(-3/2) = – 2/3

cot 210o = x/y = (-3/2)/(-1/2) = 3

sin 210o

sin (7π/6)c

cos 210o



cos (7π/6)c

tan 210o

tan (7π/6)c

cosec 210o

cosec (π/6)c

sec 210o

sec (7π/6)c

cot 210o

cot (7π/6)c



– 1/2 – √3/2 1/3 – 2 – 2/√3

3

Trigonometric Ratios of 225o or (5π/4)c:

Let m∠ AOP = θ = 225o = (5π/4)c

m∠ POM = 45o



Ray OA is the initial arm of the angle.

The terminal arm of the angle ray OP intersects the circle at P(x, y)

Let PM be perpendicular to OX’. Thus ΔOMP is 45o-45o-90o triangle

PM = 1/2(OP) = 1/2 (1) = 1/2  (side opposite to 45o)

OM = 1/2(OP) = 1/2 (1) = √1/√2  (side opposite to 45o)

Point P is in the third quadrant



Hence x = – 1/2 an y = – 1/2. Thus

sin 225o = y = – 1/2

cos 225o =  x = – 1/√2

tan 22o =  y/x  = (-1/2)/(- 1/2) =  1



cosec 225o = 1/y = 1/(-1/2) = – 2

sec 225o  =  1/x = 1/(-1/2) = – 2

cot 225o = x/y = (-1/2)/(-1/2) = 1

sin 225o

sin (5π/4)c

cos 225o

cos (5π/4)c

tan 225o



tan (5π/4)c

cosec 225o

cosec (5π/4)c

sec 225o

sec (5π/4)c

cot 225o

cot (5π/4)c

– 1/2 – 1/√2 1 – √2 – √2

1

Trigonometric Ratios of 240o or (4π/3)c:

Let m∠ AOP = θ = 240o = (4π/3)c

m∠ POM = 60o

Let PM be perpendicular to OX’. Thus ΔOMP is 30o-60o-90o triangle

PM = 3/2(OP) = 1/2 (1) = 3/2  (side opposite to 60o)

OM = 1/2(OP) = 3/2 (1) = 1/2  (side opposite to 30o)

Point P is in the third quadrant

Hence x = – 1/2 an y = – 3/2. Thus

sin 240o = y = – 3/2

cos 240o =  x = –1/2

tan 240o =  y/x  = (-3/2)/(-1/2) = 3

cosec 240o = 1/y = 1/(-3/2) = – 2/3

sec 240o  =  1/x = 1/(-1/2) = – 2

cot 240o = x/y = (-1/2)/(-3/2) = 1/3

sin 240o

sin (4π/3)c

cos 240o

cos (4π/3)c

tan 240o

tan (4π/3)c

cosec 240o

cosec (4π/3)c

sec 240o

sec (4π/3)c

cot 240o

cot (4π/3)c

– √3/2 -1/2 3 – 2/3 – 2

1/√3

Trigonometric Ratios of 270o or (3π/2)c:

Let m∠ AOP = θ = 270o = (3π/2)c

Let PM be perpendicular to OX. M coicides with O

PM = 1  and OM = 0

Point P is on negative y-axis

Hence x = 0 an y = –1. Thus

sin 270o = y = –1

cos 270o =  x = 0

tan 270o =  y/x  (Not defined since x = 0)

cosec 270o = 1/y = 1/-1 = -1

sec 270o  =  1/x  (Not defined since x = 0)

cot 270o = x/y = 0/-1 = 0

sin 270o

sin (π/2)c

cos 270o

cos (π/2)c

tan 270o

tan (π/2)c

cosec 270o

cosec (π/2)c

sec 270o

sec (π/2)c

cot 270o

cot (π/2)c

– 1 0 – 1

0

Trigonometric Ratios of 300o or (5π/3)c:

Let m∠ AOP = θ = 300o = (5π/3)c

Let PM be perpendicular to OX. Thus ΔOMP is 30o-60o-90o triangle

PM = 3/2(OP) = 1/2 (1) = 3/2  (side opposite to 60o)

OM = 1/2(OP) = 3/2 (1) = 1/2  (side opposite to 30o)

Point P is in the fourth quadrant

Hence x = 1/2 an y = – 3/2. Thus

sin 300o = y = – 3/2

cos 300o =  x = 1/2

tan 300o =  y/x  = (-3/2)/(1/2) = – 3

cosec 300o = 1/y = 1/(- 3/2) = – 2/3

sec 300o  =  1/x = 1/(1/2) = 2

cot 300o = x/y = (1/2)/(- 3/2) = – 1/3

sin 300o

sin (5π/3)c

cos 300o

cos (5π/3)c

tan 300o

tan (5π/3)c

cosec 300o

cosec (5π/3)c

sec 300o

sec (5π/3)c

cot 30o

cot (5π/3)c

– √3/2 1/2 – √3 – 2/3 2

– 1/√3

Trigonometric Ratios of 315o or (7π/4)c:

Let m∠ AOP = θ = 315o = (7π/4)c

Ray OA is the initial arm of the angle.

The terminal arm of the angle ray OP intersects the circle at P(x, y)

Let PM be perpendicular to OX. Thus ΔOMP is 45o-45o-90o triangle

PM = 1/2(OP) = 1/2 (1) = 1/2  (side opposite to 45o)

OM = 1/2(OP) = 1/2 (1) = 1/√2  (side opposite to 45o)

Point P is in the fourth quadrant

Hence x = 1/2 an y = -1/2. Thus

sin 315o = y = – 1/2

cos 315o =  x = √1/√2

tan 315o =  y/x  = (-1/2)/(1/2) = – 1

cosec 315o = 1/y = 1/(-1/2) = – 2

sec 315o  =  1/x = 1/(1/2) = 2

cot 315o = x/y = (1/2)/(-1/2) = – 1

sin 315o

sin (7π/4)c

cos 315o

cos (7π/4)c

tan 315o

tan (7π/4)c

cosec 315o

cosec (7π/4)c

sec 315o

sec (7π/4)c

cot 315o

cot (7π/4)c

– 1/2 1/√2 – 1 – √2 2

– 1

Trigonometric Ratios of 330o or (11π/6)c:

Let m∠ AOP = θ = 330o = (11π/6)c

Ray OA is the initial arm of the angle.

The terminal arm of the angle ray OP intersects the circle at P(x, y)

Let PM be perpendicular to OX. Thus ΔOMP is 30o-60o-90o triangle

PM = 1/2(OP) = 1/2 (1) = 1/2  (side opposite to 30o)

OM = 3/2(OP) = 3/2 (1) = 3/2  (side opposite to 60o)

Point P is in the fourth quadrant

Hence x = 3/2 an y = – 1/2. Thus

sin 330o = y = – 1/2

cos 330o =  x = 3/2

tan 330o =  y/x  = (-1/2)/(3/2) = – 1/3

cosec 330o = 1/y = 1/(-1/2) = – 2

sec 330o  =  1/x = 1/(3/2) = 2/3

cot 330o = x/y = (3/2)/(-1/2) = – 3

sin 330o

sin (11π/6)c

cos 330o

cos (11π/6)c

tan 330o

tan (11π/6)c

cosec 330o

cosec (11π/6)c

sec 330o

sec (11π/6)c

cot 330o

cot (11π/6)c

– 1/2 3/2 – 1/3 – 2 2/√3

– √3

Trigonometric Ratios of 360o or 2πc:

Trigonometric Ratios

Let m∠ AOP = θ = 360o = 2πc

Ray OA is the initial arm of the angle.

The terminal arm of the angle ray OP intersects the circle at P(1, 0)

Hence x = 1 an y = 0. Thus

sin 360o = y = 0

cos 360o =  x = 1

tan 360o =  y/x  = 0/1 = 0

cosec 360o = 1/y (Not defined since y = 0)

sec 360o  =  1/x = 1/1 = 1

cot 360o = x/y (Not defined since y = 0)

sin 360o

sin (π)c

cos 360o

cos (π)c

tan 360o

tan (π)c

cosec 360o

cosec (π)c

sec 360o

sec (π)c

cot 360o

cot (π)c

0 1 0 1

 

Science > Mathematics > TrigonometryYou are Here
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