# TRIGONOMETRIC RATIOS OF SUPPLEMENTARY ANGLES

Two angles are supplementary to each other if their sum is equal to 180°

Trigonometric-ratios of supplementary angles are given below.

 sin (180° - θ)  =  sin θcos (180° - θ)  =  - cos θtan (180° - θ)  =  - tan θcsc (180° - θ)  =  csc θsec (180° - θ)  =  - sec θcot (180° - θ)  =  - cot θ sin (180° + θ)  =  - sin θcos (180° + θ)  =  - cos θtan (180° + θ)  =  tan θcsc (180° + θ)  =  - csc θsec (180° + θ)  =  - sec θcot (180° + θ)  =  cot θ

Let us see, how the trigonometric ratios of supplementary 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 formula has been explained clearly in the figure given below. More clearly From the above picture, it is very clear that

(i)  (180° - θ) falls in the second quadrant and

(i)  (180° + θ) falls in the third quadrant

In the second quadrant (180° - θ), sin and csc are positive and other trigonometric ratios are negative.

In the third quadrant (180° + θ), 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 Using ASTC Formula

Example 1 :

Evaluate :

sin (180° - θ)

Solution :

To evaluate sin (180° - θ), we have to consider the following important points.

(i)  (180° - θ) will fall in the IInd quadrant.

(ii)  When we have 180°, "sin" will not be changed as "cos"

(iii)  In the IInd quadrant, the sign of "sin" is positive.

Considering the above points, we have

sin (180° - θ)  =  sin θ

Example 2 :

Evaluate :

cos (180° - θ)

Solution :

To evaluate cos (180° - θ), we have to consider the following important points.

(i)  (180° - θ) will fall in the IInd quadrant.

(ii)  When we have 180°, "cos" will not be changed as "sin"

(iii)  In the IInd quadrant, the sign of "cos" is negative.

Considering the above points, we have

cos (180° - θ)  =  - cos θ

Example 3 :

Evaluate :

tan (180° - θ)

Solution :

To evaluate tan (180° - θ), we have to consider the following important points.

(i)  (180° - θ) will fall in the IInd quadrant.

(ii)  When we have 180°, "tan" will not be changed as "cot"

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

Considering the above points, we have

tan (180° - θ)  =  - tan θ

Example 4 :

Evaluate :

csc (180° - θ)

Solution :

To evaluate csc (180° - θ), we have to consider the following important points.

(i)  (180° - θ) will fall in the IInd quadrant.

(ii)  When we have 180°, "csc" will not be changed as "sec"

(iii)  In the IInd quadrant, the sign of "csc" is positive.

Considering the above points, we have

csc (180° - θ)  =  csc θ

Example 5 :

Evaluate :

sec (180° - θ)

Solution :

To evaluate sec (180° - θ), we have to consider the following important points.

(i)  (180° - θ) will fall in the IInd quadrant.

(ii)  When we have 180°, "sec" will not be changed as "csc"

(iii)  In the IInd quadrant, the sign of "sec" is negative.

Considering the above points, we have

sec (180° - θ)  =  - sec θ

Example 6 :

Evaluate :

cot (180° - θ)

Solution :

To evaluate cot (180° - θ), we have to consider the following important points.

(i)  (180° - θ) will fall in the IInd quadrant.

(ii)  When we have 180°, "cot" will not be changed as "tan"

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

Considering the above points, we have

cot (180° - θ)  =  - cot θ

Example 7 :

Evaluate :

sin (180° + θ)

Solution :

To evaluate sin (180° + θ), we have to consider the following important points.

(i)  (180° + θ) will fall in the IIIrd quadrant.

(ii)  When we have 180°, "sin" will not be changed as "cos"

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

Considering the above points, we have

sin (180° + θ)  =  - sin θ

Example 8 :

Evaluate :

cos (180° + θ)

Solution :

To evaluate cos (180° + θ), we have to consider the following important points.

(i)  (180° + θ) will fall in the IIIrd quadrant.

(ii)  When we have 180°, "cos" will not be changed as "sin"

(iii)  In the IIIrd quadrant, the sign of "cos" is negative.

Considering the above points, we have

cos (180° + θ)  =  - cos θ

Example 9 :

Evaluate :

tan (180° + θ)

Solution :

To evaluate tan (180° + θ), we have to consider the following important points.

(i)  (180° + θ) will fall in the IIIrd quadrant.

(ii)  When we have 180°, "tan" will not be changed as "cot"

(iii)  In the IIIrd quadrant, the sign of "tan" is positive.

Considering the above points, we have

tan (180° + θ)  =  tan θ

Example 10 :

Evaluate :

csc (180° + θ)

Solution :

To evaluate csc (180° + θ), we have to consider the following important points.

(i)  (180° + θ) will fall in the IIIrd quadrant.

(ii)  When we have 180°, "csc" will not be changed as "sec"

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

Considering the above points, we have

csc (180° + θ)  =   - csc θ

Example 11 :

Evaluate :

sec (180° + θ)

Solution :

To evaluate sec (180° + θ), we have to consider the following important points.

(i)  (180° + θ) will fall in the IIIrd quadrant.

(ii)  When we have 180°, "sec" will not be changed as "csc"

(iii)  In the IIIrd quadrant, the sign of "sec" is negative.

Considering the above points, we have

sec (180° + θ)  =  - sec θ

Example 12 :

Evaluate :

cot (180° + θ)

Solution :

To evaluate cot (180° + θ), we have to consider the following important points.

(i)  (180° + θ) will fall in the IIIrd quadrant.

(ii)  When we have 180°, "cot" will not be changed as "tan"

(iii)  In the IIIrd quadrant, the sign of "cot" is positive.

Considering the above points, we have

cot (180° + θ)  =  cot θ

## Summary (Supplementary Angles)

 sin (180° - θ)  =  sin θcos (180° - θ)  =  - cos θtan (180° - θ)  =  - tan θcsc (180° - θ)  =  csc θsec (180° - θ)  =  - sec θcot (180° - θ)  =  - cot θ sin (180° + θ)  =  - sin θcos (180° + θ)  =  - cos θtan (180° + θ)  =  tan θcsc (180° + θ)  =  - csc θsec (180° + θ)  =  - sec θcot (180° + θ)  =  cot θ

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