What are the relations among all the trigonometrical ratios of (270° + θ)?
In trigonometrical ratios of angles (270° + 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 (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 |
Using the above proved results we will prove all six trigonometrical ratios of (180° - θ).
sin (270° + θ) = sin [1800 + 90° + θ]
= sin [1800 + (90° + θ)]
= - sin (90° + θ), [since sin (180° + θ) = - sin θ]
Therefore, sin (270° + θ) = - cos θ, [since sin (90° + θ) = cos θ]
cos (270° + θ) = cos [1800 + 90° + θ]
= cos [I 800 + (90° + θ)]
= - cos (90° + θ), [since cos (180° + θ) = - cos θ]
Therefore, cos (270° + θ) = sin θ, [since cos (90° + θ) = - sin θ]
tan ( 270° + θ) = tan [1800 + 90° + θ]
= tan [180° + (90° + θ)]
= tan (90° + θ), [since tan (180° + θ) = tan θ]
Therefore, tan (270° + θ) = - cot θ, [since tan (90° + θ) = - cot θ]
csc (270° + θ) = \(\frac{1}{sin (270° + \Theta)}\)
= \(\frac{1}{- cos \Theta}\), [since sin (270° + θ) = - cos θ]
Therefore, csc (270° + θ) = - sec θ;
sec (270° + θ) =\(\frac{1}{cos (270° + \Theta)}\)
= \(\frac{1}{sin \Theta}\), [since cos (270° + θ) = sin θ]
Therefore, sec (270° + θ) = csc θ
and
cot (270° + θ) = \(\frac{1}{tan (270° + \Theta)}\)
= \(\frac{1}{- cot \Theta}\), [since tan (270° + θ) = - cot θ]
Therefore, cot (270° + θ) = - tan θ.
Solved examples:
1. Find the value of csc 315°.
Solution:
csc 315° = sec (270 + 45)°
= - sec 45°; since we know, csc (270° + θ) = - sec θ
= - √2
2. Find the value of cos 330°.
Solution:
cos 330° = cos (270 + 60)°
= sin 60°; since we know, cos (270° + θ) = sin θ
= \(\frac{√3}{2}\)
● Trigonometric Functions
11 and 12 Grade Math
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