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

• Basic Trigonometric Ratios and Their Names
• Restrictions of Trigonometrical Ratios
• Reciprocal Relations of Trigonometric Ratios
• Quotient Relations of Trigonometric Ratios
  
• Limit of Trigonometric Ratios
  
• Trigonometrical Identity
• Problems on Trigonometric Identities
• Elimination of Trigonometric Ratios
• Eliminate Theta between the equations
• Problems on Eliminate Theta
• Trig Ratio Problems
• Proving Trigonometric Ratios
• Trig Ratios Proving Problems
• Verify Trigonometric Identities
• Trigonometrical Ratios of 0°
• Trigonometrical Ratios of 30°
• Trigonometrical Ratios of 45°
• Trigonometrical Ratios of 60°
• Trigonometrical Ratios of 90°
• Trigonometrical Ratios Table
• Problems on Trigonometric Ratio of Standard Angle
  
• Trigonometrical Ratios of Complementary Angles
• Rules of Trigonometric Signs
• Signs of Trigonometrical Ratios
  
• All Sin Tan Cos Rule
• Trigonometrical Ratios of (- θ)
• Trigonometrical Ratios of (90° + θ)
• Trigonometrical Ratios of (90° - θ)
• Trigonometrical Ratios of (180° + θ)
• Trigonometrical Ratios of (180° - θ)
• Trigonometrical Ratios of (270° + θ)
• Trigonometrical Ratios of (270° - θ)
• Trigonometrical Ratios of (360° + θ)
• Trigonometrical Ratios of (360° - θ)
• Trigonometrical Ratios of any Angle
• Trigonometrical Ratios of some Particular Angles
• Trigonometric Ratios of an Angle
• Trigonometric Functions of any Angles
• Problems on Trigonometric Ratios of an Angle
• Problems on Signs of Trigonometrical Ratios

11 and 12 Grade Math

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