What are the relations among all the trigonometrical ratios of (180° - θ)?
In trigonometrical ratios of angles (180° - θ) we will find the relation between all six trigonometrical ratios.
We know that, 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 |
and 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 |
Using the above proved results we will prove all six trigonometrical ratios of (180° - θ).
sin (180° - θ) = sin (90° + 90° - θ)
= sin [90° + (90° - θ)]
= cos (90° - θ), [since sin (90° + θ) = cos θ]
Therefore, sin (180° - θ) = sin θ, [since cos (90° - θ) = sin θ]
cos (180° - θ) = cos (90° + 90° - θ)
= cos [90° + (90° - θ)]
= - sin (90° - θ), [since cos (90° + θ) = -sin θ]
Therefore, cos (180° - θ) = - cos θ, [since sin (90° - θ) = cos θ]
tan (180° - θ) = cos (90° + 90° - θ)
= tan [90° + (90° - θ)]
= - cot (90° - θ), [since tan (90° + θ) = -cot θ]
Therefore, tan (180° - θ) = - tan θ, [since cot (90° - θ) = tan θ]
csc (180° - θ) = \(\frac{1}{sin (180° - \Theta)}\)
= \(\frac{1}{sin \Theta}\), [since sin (180° - θ) = sin θ]
Therefore, csc (180° - θ) = csc θ;
sec (180° - θ) = \(\frac{1}{cos (180° - \Theta)}\)
= \(\frac{1}{- cos \Theta}\), [since cos (180° - θ) = - cos θ]
Therefore, sec (180° - θ) = - sec θ
and
cot (180° - θ) = \(\frac{1}{tan (180° - \Theta)}\)
= \(\frac{1}{- tan \Theta}\), [since tan (180° - θ) = - tan θ]
Therefore, cot (180° - θ) = - cot θ.
Solved examples:
1. Find the value of sec 150°.
Solution:
sec 150° = sec (180 - 30)°
= - sec 30°; since we know, sec (180° - θ) = - sec θ
= - \(\frac{2}{√3}\)
2. Find the value of tan 120°.
Solution:
tan 120° = tan (180 - 60)°
= - tan 60°; since we know, tan (180° - θ) = - tan θ
= - √3
● Trigonometric Functions
11 and 12 Grade Math
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