TRIGONOMETRIC RATIOS OF 270 DEGREE MINUS THETA 

Trigonometric ratios of 270 degree minus theta is one of the branches of ASTC formula in trigonometry. 

Trigonometric-ratios of 270 degree minus theta are given below.

sin (270° - θ)  =  - cos θ

cos (270° - θ)  =  - sin θ

tan (270° - θ)  =  cot θ

csc (270° - θ)  =  - sec θ

sec (270° - θ)  =  - cos θ

cot (270° - θ)  =  tan θ

Let us see, how the trigonometric ratios of 270 degree minus theta are determined. 

To know that, first we have to understand ASTC formula. 

The ASTC formula can be remembered easily using the following phrases.

"All Sliver Tea Cups" 

or

"All Students Take Calculus"

ASTC formula has been explained clearly in the figure given below.

More clearly 

From the above picture, it is very clear that the angle 270 degree minus theta falls in the third quadrant. 

In the third quadrant (270° degree minus theta), tan and cot are positive and other trigonometric ratios are negative.

Important Conversions

When we have the angles 90° and 270° in the trigonometric ratios in the form of

(90° + θ)

(90° - θ)

(270° + θ)

(270° - θ)

We have to do the following conversions, 

sin θ <------> cos θ

tan θ <------> cot θ

csc θ <------> sec θ

For example,

sin (270° + θ)  =  - cos θ

cos (90° - θ)  =  sin θ

For the angles 0° or 360° and  180°, we should not make the above conversions. 

Evaluation of Trigonometric Ratios 270 Degree Minus Theta

Problem 1 :

Evaluate :

sin (270° - θ)

Solution :

To evaluate sin (270° - θ), we have to consider the following important points. 

(i)  (270° - θ) will fall in the III^rd quadrant. 

(ii)  When we have 270°, "sin" will become "cos"

(iii)  In the III^rd quadrant, the sign of "sin" is negative. 

Considering the above points, we have 

sin (270° - θ)  =  - cos θ

Problem 2 :

Evaluate :

cos (270° - θ)

Solution :

To evaluate cos (270° - θ), we have to consider the following important points. 

(i)  (270° - θ) will fall in the III^rd quadrant. 

(ii)  When we have 270°, "cos" will become "sin"

(iii)  In the III^rd quadrant, the sign of "cos" is negative. 

Considering the above points, we have 

cos (270° - θ)  =  - sin θ

Problem 3 :

Evaluate :

tan (270° - θ)

Solution :

To evaluate tan (270° - θ), we have to consider the following important points. 

(i)  (270° - θ) will fall in the III^rd quadrant. 

(ii)  When we have 270°, "tan" will become "cot"

(iii)  In the III^rd quadrant, the sign of "tan" is positive. 

Considering the above points, we have 

tan (270° - θ)  =  cot θ

Problem 4 :

Evaluate :

csc (270° - θ)

Solution :

To evaluate csc (270° - θ), we have to consider the following important points. 

(i)  (270° - θ) will fall in the III^rd quadrant. 

(ii)  When we have 270°, "csc" will become "sec"

(iii)  In the III^rd quadrant, the sign of "csc" is negative. 

Considering the above points, we have 

csc (270° - θ)  =  - sec θ

Problem 5 :

Evaluate :

sec (270° - θ)

Solution :

To evaluate sec (270° - θ), we have to consider the following important points. 

(i)  (270° - θ) will fall in the III^rd quadrant. 

(ii)  When we have 270°, "sec" will become "csc"

(iii)  In the III^rd quadrant, the sign of "sec" is negative. 

Considering the above points, we have 

sec (270° - θ)  =  - csc θ

Problem 6 :

Evaluate :

cot (270° - θ)

Solution :

To evaluate cot (270° - θ), we have to consider the following important points. 

(i)  (270° - θ) will fall in the III^rd quadrant. 

(ii)  When we have 270°, "cot" will become "tan"

(iii)  In the III^rd quadrant, the sign of "cot" is positive. 

Considering the above points, we have 

cot (270° - θ)  =  tan θ

Summary (270 Degree - θ)

sin (270° - θ)  =  - cos θ

cos (270° - θ)  =  - sin θ

tan (270° - θ)  =  cot θ

csc (270° - θ)  =  - sec θ

sec (270° - θ)  =  - cos θ

cot (270° - θ)  =  tan θ

Evaluate the six trigonometric functions of θ.

Problem 7 :

Solution :

Opposite side = -3

Adjacent side = 4

Hypotenuse = √(-3)² + 4²

= √(9 + 16)

= √25

= 5

Since the shown angle lies in the 4th quadrant, for the trigonometric ratios cos θ and sec θ. will be positive.

sin θ = Opposite side / hypotenuse

sin θ = -3/5

cos θ = Adjacent side / hypotenuse

cos θ = 4/5

tan θ = Opposite side / Adjacent side

tan θ = -3/4

csc θ = hypotenuse/Opposite side

csc θ = -5/3

sec θ = hypotenuse/Adjacent side

sec θ = 5/4

cot θ = Adjacent side / Opposite side

cot θ = -4/3 

Problem 8 :

Solution :

Opposite side = -12

Adjacent side = 5

Hypotenuse = √5² + (-12)²

= √(25 + 144)

= √169

= 13

Since the shown angle lies in the 4th quadrant, for the trigonometric ratios cos θ and sec θ. will be positive.

sin θ = -12/13

cos θ = 5/13

tan θ = -12/5

csc θ = -13/12

sec θ = 13/5

cot θ = -5/12 

Problem 8 :

Evaluate the function without using a calculator.

a)  sec 135°

b)  tan 240°

c) sin(−150°)

d) csc(−420°)

Solution :

a)  sec 135°

Since the angle lies in the second quadrant, reference angle = 180 - θ

= 180 - 135

= 45

sec 135 = sec 45

= 1/cos 45

= 1/(1/√2)

= √2

b)  tan 240°

Since the angle lies in the third quadrant, reference angle = θ - 180

= 240 - 180

= 60

tan 240 = tan 60

= √3

c) sin(−150°) = - sin 150

Since the angle lies in the second quadrant, reference angle = 180 - θ

= 180 - 150

= 30

- sin 150 = - sin 30

= -1/2

d) csc(−420°) = - csc 420

= -csc 60

= -1/sin 60

= -1/(√3/2)

= -2/√3

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