Sign of Trigonometric Function :
To find the sign of trigonometric function, we should know about the four quadrants.
When we have the angles 90° and 270° in the trigonometric ratios in the form of
(90° + θ) (or) (90° - θ)
(270° + θ) (or) (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.
Ist |
A |
All 6 trigonometric ratios are positive |
IInd |
S |
Sin θ and its reciprocal cosec θ are positive. The other trigonometric ratios are negative |
IIIrd |
T |
tan θ and its reciprocal cot θ are positive. The other trigonometric ratios are negative |
IVth |
C |
cos θ and its reciprocal sec θ are positive. The other trigonometric ratios are negative |
Example 1 :
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
Example 2 :
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 III rd quadrant.
(ii) When we have 180°, "sin" will not be changed
(iii) In the III rd quadrant, the sign of "sin" is negative.
Considering the above points, we have
sin (180° + θ) = - sin θ
Based on the above two examples, we can evaluate the following trigonometric ratios.
-θ |
90° - θ |
sin (-θ) = - sin θ cos (-θ) = cos θ tan (-θ) = - tan θ csc (-θ) = - csc θ sec (-θ) = sec θ cot (-θ) = - cot θ |
sin (90°-θ) = cos θ cos (90°-θ) = sin θ tan (90°-θ) = cot θ csc (90°-θ) = sec θ sec (90°-θ) = csc θ cot (90°-θ) = tan θ |
90° + θ |
180° - θ |
sin (90°+θ) = cos θ cos (90°+θ) = -sin θ tan (90°+θ) = -cot θ csc (90°+θ) = sec θ sec (90°+θ) = -csc θ cot (90°+θ) = -tan θ |
sin (180°-θ) = sin θ cos (180°-θ) = -cos θ tan (180°-θ) = -tan θ csc (180°-θ) = csc θ sec (180°-θ) = -sec θ cot (180°-θ) = -cot θ |
180° + θ |
270° - θ |
sin (180°+θ) = -sin θ cos (180°+θ) = -cos θ tan (180°+θ) = tan θ csc (180°+θ) = -csc θ sec (180°+θ) = -sec θ csc (180°+θ) = cot θ |
sin (270°-θ) = -cos θ cos (270°-θ) = -sin θ tan (270°-θ) = cot θ csc (270°-θ) = -sec θ sec (270°-θ) = -csc θ cot (270°-θ) = tan θ |
270° + 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
After having gone through the stuff given above, we hope that the students would have understood how to find the sign of a trigonometric function.
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