TRIGONOMETRIC RATIOS OF NEGATIVE ANGLES

Trigonometric ratios of negative angles is a part of ASTC concept in trigonometry. 

The results of trigonometric ratios for negative angles are given below.

sin (- θ)  =  - sin θ

cos (- θ)  =  cos θ

tan (- θ)  =  - tan θ

csc (- θ)  =  -csc θ

sec (- θ)  =  sec θ

cot (- θ)  =  - cot θ

Let us see, how the trigonometric ratios of negative angles 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 formla has been explained clearly in the figure given below.

More clearly 

Key Concept - Negative angles

Whenever we have negative angles in trigonometric ratios, we have to assume that it falls in the fourth quadrant.

In the fourth quadrant ( - θ or 360° - θ ), cos and sec 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 of Negative Angles

Problem 1 :

Evaluate :

sin (- θ)

Solution :

sin (- θ)  can be written as sin (0° - θ)

That is,  sin (- θ)  =  sin (0° - θ)

To evaluate sin (0° - θ), we have to consider the following important points. 

(i)  (0° - θ) will fall in the IV th quadrant. 

(ii)  When we have 0°, "sin" will not be changed as "cos"

(iii)  In the IV th quadrant, the sign of "sin" is negative. 

Considering the above points, we have 

sin (- θ)  =  sin (0° - θ)  =   - sin θ

Problem 2 :

Evaluate :

cos (- θ)

Solution :

cos (- θ)  can be written as cos (0° - θ)

That is,  cos (- θ)  =  cos (0° - θ)

To evaluate cos (0° - θ), we have to consider the following important points. 

(i)  (0° - θ) will fall in the IV th quadrant. 

(ii)  When we have 0°, "cos" will not be changed as "sin"

(iii)  In the IV th quadrant, the sign of "cos" is positive. 

Considering the above points, we have 

cos (- θ)  =  cos (0° - θ)  =   cos θ

Problem 3 :

Evaluate :

tan (- θ)

Solution :

tan (- θ)  can be written as tan (0° - θ)

That is,  tan (- θ)  =  tan (0° - θ)

To evaluate tan (0° - θ), we have to consider the following important points. 

(i)  (0° - θ) will fall in the IV th quadrant. 

(ii)  When we have 0°, "tan" will not be changed as "cot"

(iii)  In the IV th quadrant, the sign of "tan" is negative. 

Considering the above points, we have 

tan (- θ)  =  tan (0° - θ)  =   - tan θ

Problem 4 :

Evaluate :

csc (- θ)

Solution :

csc (- θ)  can be written as csc (0° - θ)

That is,  csc (- θ)  =  csc (0° - θ)

To evaluate csc (0° - θ), we have to consider the following important points. 

(i)  (0° - θ) will fall in the IV th quadrant. 

(ii)  When we have 0°, "csc" will not be changed as "sec"

(iii)  In the IV th quadrant, the sign of "csc" is negative. 

Considering the above points, we have 

csc (- θ)  =  csc (0° - θ)  =   - csc θ

Problem 5 :

Evaluate :

sec (- θ)

Solution :

sec (- θ)  can be written as sec (0° - θ)

That is,  sec (- θ)  =  sec (0° - θ)

To evaluate sec (0° - θ), we have to consider the following important points. 

(i)  (0° - θ) will fall in the IV th quadrant. 

(ii)  When we have 0°, "sec" will not be changed as "csc"

(iii)  In the IV th quadrant, the sign of "sec" is positive. 

Considering the above points, we have 

sec (- θ)  =  sec (0° - θ)  =   sec θ

Problem 6 :

Evaluate :

cot (- θ)

Solution :

cot (- θ)  can be written as cot (0° - θ)

That is,  cot (- θ)  =  cot (0° - θ)

To evaluate cot (0° - θ), we have to consider the following important points. 

(i)  (0° - θ) will fall in the IV th quadrant. 

(ii)  When we have 0°, "cot" will not be changed as "tan"

(iii)  In the IV th quadrant, the sign of "cot" is negative. 

Considering the above points, we have 

cot (- θ)  =  cot (0° - θ)  =   - cot θ

Summary (Negative Angles)

sin (- θ)  =  - sin θ

cos (- θ)  =  cos θ

tan (- θ)  =  - tan θ

csc (- θ)  =  -csc θ

sec (- θ)  =  sec θ

cot (- θ)  =  - cot θ

Angles Equal to 360°

If the angle is equal to or greater than 360°, we have to divide the given angle by 360 and take the remainder. 

For example,  

(i) Let us consider the angle 450°.

When we divide 450° by 360, we get the remainder 90°. 

Therefore, 450°  =  90°

(ii) Let us consider the angle 360°

When we divide 360° by 360, we get the remainder 0°.

Therefore, 360°  =  0°

Based on the above two examples, we can evaluate the following trigonometric ratios. 

sin (360° - θ)  =  sin (0° - θ)  =  sin (- θ)  =  - sin θ

cos (360° - θ)  =  cos (0° - θ)  =  cos (- θ)  =  cos θ

tan (360° - θ)  =  tan (0° - θ)  =  tan (- θ)  =  - tan θ

csc (360° - θ)  =  csc (0° - θ)  =  csc (- θ)  =  - csc θ

sec (360° - θ)  =  sec (0° - θ)  =  sec (- θ)  =  sec θ

cot (360° - θ)  =  cot (0° - θ)  =  cot (- θ)  =  - cot θ

After having gone through the stuff given above, we hope that the students would have understood how to evaluate trigonometric ratios of negative angles.

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