# Trigonometrical Ratios of (180° + θ)

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

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 example:

1. Find the value of sin 225°.

Solution:

sin (225)° = sin (180 + 45)°

= - sin 45°; since we know sin (180° + θ) = - sin θ

= - $$\frac{1}{√2}$$

2. Find the value of sec 210°.

Solution:

sec (210)° = sec (180 + 30)°

= - sec 30°; since we know sec (180° + θ) = - sec θ

= - $$\frac{1}{√2}$$

3. Find the value of tan 240°.

Solution:

tan (240)° = tan (180 + 60)°

= tan 60°; since we know tan (180° + θ) = tan θ

= √3

Trigonometric Functions

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