Trigonometrical Ratios of (270° + θ)

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Trigonometrical Ratios of (270° + θ)

2023-10-01 02:37| 来源: 网络整理| 查看: 265

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

Basic Trigonometric Ratios and Their NamesRestrictions of Trigonometrical RatiosReciprocal Relations of Trigonometric RatiosQuotient Relations of Trigonometric RatiosLimit of Trigonometric RatiosTrigonometrical IdentityProblems on Trigonometric IdentitiesElimination of Trigonometric RatiosEliminate Theta between the equationsProblems on Eliminate ThetaTrig Ratio ProblemsProving Trigonometric RatiosTrig Ratios Proving ProblemsVerify Trigonometric IdentitiesTrigonometrical Ratios of 0°Trigonometrical Ratios of 30°Trigonometrical Ratios of 45°Trigonometrical Ratios of 60°Trigonometrical Ratios of 90°Trigonometrical Ratios TableProblems on Trigonometric Ratio of Standard AngleTrigonometrical Ratios of Complementary AnglesRules of Trigonometric SignsSigns of Trigonometrical RatiosAll Sin Tan Cos RuleTrigonometrical 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 AngleTrigonometrical Ratios of some Particular AnglesTrigonometric Ratios of an AngleTrigonometric Functions of any AnglesProblems on Trigonometric Ratios of an AngleProblems on Signs of Trigonometrical Ratios

11 and 12 Grade Math

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