Trigonometrical Ratios of (- θ)

What is the relation among all the trigonometrical ratios of (– θ)?

In trigonometrical ratios of angles (- θ) we will find the relation between all six trigonometrical ratios.

Let a rotating line OA rotates about O in the anti-clockwise direction. From initial position to ending position OA make an angle ∠XOA = θ.

Trigonometrical Ratios of (- θ)

Diagram 1

Trigonometrical Ratios of (- θ)

Diagram 2

Again a rotating line OA rotates about O in the clockwise direction and makes an angle ∠XOB having magnitude equal to ∠XOA.

Then we get, ∠XOB = - θ. Observe the diagram 1 and 4 to take a point C on OA and draw CD perpendicular to OX. Or we can also observe the diagram 2 and 3 where CD perpendicular to OX'. Let produce CD to intersect OB at E. Now, from the ∆ COD and ∆ EOD we get ∠COD = ∠EOD (same magnitude), ∠ODC = ∠ODE and OD is common.

Therefore, ∆ COD ≅ ∆ EOD (congruent)

Therefore,  according to the rules of trigonometric sign we get,

ED = - CD and OE = OC.

Again according to the definition of trigonometric ratios,

sin (- θ) = \(\frac{ED}{OE}\)

sin (- θ) = \(\frac{- CD}{OC}\), [ED = CD and OE = OC since, ∆ COD ≅ ∆ EOD]

sin (- θ) = - sin θ


again, cos (- θ) = \(\frac{OD}{OE}\)

cos (- θ) = \(\frac{OD}{OC}\), [OE = OC since, ∆ COD ≅ ∆ EOD]

cos (- θ) = cos θ


again, tan (- θ) = \(\frac{ED}{OD}\)

tan (- θ) = \(\frac{- CD}{OD}\), [ED = CD since, ∆ COD ≅ ∆ EOD]

tan (- θ) = -  tan θ.


similarly, csc (- θ) = \(\frac{1}{sin (- \Theta)}\)

csc (- θ) = \(\frac{1}{-  sin \Theta}\)

csc (- θ) = - csc θ.


again, sec (- θ) = \(\frac{1}{cos (- \Theta)}\)

sec (- θ) = \(\frac{1}{cos \Theta}\) 

sec (- θ) = sec θ.


And again, cot (- θ) = \(\frac{1}{tan (- \Theta)}\)

cot (- θ) = \(\frac{1}{-  tan \Theta}\)

cot (- θ) = - cot θ.


Solved example:

1. Find the value of sin (- 45)°.

Solution:

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

               = \(\frac{-1}{√2}\)


2. Find the value of sec (- 60)°.

Solution:

sec (- 60)° = sec 60°; since we know sec (- θ) = sec θ

                = 2   


3. Find the value of cot (- 90)°.

Solution:

cot (- 90)° = - tan 90°; since we know cot (- θ) = - tan θ

                = 0


11 and 12 Grade Math

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