what is sin90+x and cos 1808+x
Answers
Answer:
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
and
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) = 1sin(180°−Θ)
= 1sinΘ, [since sin (180° - θ) = sin θ]
Therefore, csc (180° - θ) = csc θ;
sec (180° - θ) = 1cos(180°−Θ)
= 1−cosΘ, [since cos (180° - θ) = - cos θ]
Therefore, sec (180° - θ) = - sec θ
and
cot (180° - θ) = 1tan(180°−Θ)
= 1−tanΘ, [since tan (180° - θ) = - tan θ]
Therefore, cot (180° - θ) = - cot θ.
Solved examples:
1. Find the value of sec 150°.
Solution:
sec 150° = sec (180 - 30)°
= - sec 30°;