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
11 and 12 Grade Math
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