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.
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
● Trigonometric Functions
11 and 12 Grade Math
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