Symmetry of trigonometric functions

Extraction of formulae from the graphs

After the preceding proofs of the properties of symmetry of the curves of the trigonometric functions, you can now extract these again from the graphs. If you have produced several patterns, you can recover with their help the earlier formulae. Since for the sin, tan and cot curves the origin is a centre of symmetry, for the cos curve the ordinate axis is an axis of symmetry, you obtain important relations between positive and negative angles:

The parallel displacement is easily achieved by reflection in AB [FG = F'G] and a reflection in the ordinate axis [F'G' = C'D'], etc.)

The shear reflections in 2) and 3) are obtained by a reflection in AB [GH = G'H'] and a rotation about O [G'H' = -PR].

Geometrical properties: The sin curve has arbitrarily many centres. C is one of them. All diameters are bisected by it: PC = CQ, JC = CH. Moreover, is equal to and parallel to , the area bounded by the curve and the chord is congruent in the same sense to the area, bounded by of the curve and the chord . The curve has any number of central lines. AB is one of them. The area OEA, bounded by OA, AE and the curve OE, is oppositely oriented congruent to the area CE'A, etc.

The tan and cot curves yield corresponding formulae of symmetry.

Note that all formulae are valid for any angle a.

Since certain relations reoccur frequently, you can recall them also by the following mechanism.

Mechanical rule for frequently used relations

If one denotes the trigonometric functions by f(x) and their co-functions by cf(x), you have always for their absolute values

I. |f(180º a)| = |f(a180º)| = |f(360ºa)| = |f(a360º)| = ···|f(a)|.

At 180º and 360º, the trigonometric function is conserved.

II. |f(90ºa)| = |f(a90º)| = |f(270ºa)| = |f(a270º)| = ···|cf(a)|.

At 90º and 270º, the function becomes its cofunction.

You determine the leading sign afterwards! Since these formulae are valid for any angles, you can obtain the missing signs by finding it for a sharp angle.

You have |sin (180º + a)| = |sin a|. sin a ( I. quadrant) is positive, sin (180º+a) (III. quadrant) negative, whence the required formula is sin (180º+a) = - sina.

|tan (270º + a)| = |cot a| (I. quadrant) is positive, tan(270º + a) = (IV. quadrant) negative, whence tan(270º + a) = -cot a is valid for any angle a.

You can now use this mechanism for the reduction of any angle to a sharp angle and thus use the trigonometric tables for arbitrary angles.

Task: Find sin 135º.

|sin 135º|	=	|sin (180º - 135º)	=	sin 45º	=
+				+
		sin 135º =

You could also have gone via the co-function; however, it is an advantage to stick to one group of formulae, the first in this section. Since, in addition, the functions are now reduced to sharp angles and they have in the first quadrant always the + sign, you only need now to determine from the starting value the leading sign.

Task: Find cos 210º.

Task: Find tan 320º.

Task: Check: sin 314º = -0.7193, cos 140º = -0.7660, tan 217º = + 0.7536, cot 114º = -0.4452.

Determination of angles from trigonometric values.

Task: sin a = 1/2. How big is a?

sin is positive in the first two quadrants, whence two angles between 0º and 180º satisfy this equation. The sharp angle is a₁ = 30º. The obtuse angle follows from sin(180º - 30º) = +sin 30º, that is a₂= 150º.

Since after a complete positive or negative rotation by 360º the values of the function are repeated, two groups of angles satisfy the equation sin a = 1/2:

a₁ = 30º + k·360º and a₂= 150º + k·360º.

where k is an arbitrary positive or negative integer including 0.

The following figure presents the graphical solution of the task:

Task: Find the angles which satisfy the equation tan a = -1.192.

tan a is negative in the quadrants III and IV. The given value yields from the table j = 50º.

Consult for the solution of this task also the following graph!

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