Trigonometrical Ratios of (270° + θ)

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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