They are Pythagorean identities all trigonometric equations that hold for any value of the angle and are based on the Pythagorean theorem. The most famous of the Pythagorean identities is the fundamental trigonometric identity:
Sentwo(α) + Costwo(α) = 1
Next in importance and I use the Pythagorean identity of the tangent and secant:
Sotwo(α) + 1 = Sectwo(α)
And the Pythagorean trigonometric identity involving the cotangent and the cosecant:
1 + Ctgtwo(α) = Csctwo(α)
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The trigonometric ratios breast Y cosine they are represented in a circle of radius one (1) known as a trigonometric circle. Said circle has center at the origin of coordinates O.
The angles are measured from the positive semi-axis of the X, for example the angle α in figure 2 (see below). Counterclockwise if the angle is positive, and clockwise if it is a negative angle.
The ray with origin O and angle α is drawn, which intercepts the unit circle at point P. Point P is projected orthogonally on the horizontal axis X giving rise to point C. Similarly P is projected perpendicularly on the vertical axis Y giving place to point S.
We have the right triangle OCP at C.
It should be remembered that the trigonometric ratio breast is defined on a right triangle as follows:
The sine of an angle of the triangle is the ratio or quotient between the leg opposite the angle and the hypotenuse of the triangle.
Applied to the triangle OCP of figure 2 it would look like this:
Sen (α) = CP / OP
but CP = OS and OP = 1, so that:
Sen (α) = OS
This means that the projection OS on the Y axis has a value equal to the sine of the displayed angle. It should be noted that the maximum value of the sine of an angle (+1) occurs when α = 90º and the minimum (-1) when α = -90º or α = 270º.
Similarly, the cosine of an angle is the quotient between the leg adjacent to the angle and the hypotenuse of the triangle..
Applied to the triangle OCP of figure 2 it would look like this:
Cos (α) = OC / OP
but OP = 1, so that:
Cos (α) = OC
This means that the projection OC on the X axis has a value equal to the sine of the angle shown. It should be noted that the maximum cosine value (+1) occurs when α = 0º or α = 360º, while the minimum cosine value is (-1) when α = 180º.
For the right triangle OCP in C, the Pythagorean theorem is applied, which states that the sum of the square of the legs is equal to the square of the hypotenuse:
CPtwo + OCtwo = OPtwo
But it has already been said that CP = OS = Sen (α), that OC = Cos (α) and that OP = 1, so the previous expression can be rewritten as a function of the sine and cosine of the angle:
Sentwo(α) + Costwo(α) = 1
Just as the X axis in the trigonometric circle is the cosine axis and the Y axis the sine axis, in the same way there is the tangent axis (see figure 3) which is precisely the tangent line to the unit circle at the point B of coordinates (1, 0).
If you want to know the value of the tangent of an angle, draw the angle from the positive semi-axis of the X, the intersection of the angle with the axis of the tangent defines a point Q, the length of the segment OQ is the tangent of the angle.
This is because by definition, the tangent of angle α is the opposite leg QB between the adjacent leg OB. That is, Tan (α) = QB / OB = QB / 1 = QB.
The Pythagorean identity of the tangent can be proved by considering the right triangle OBQ at B (Figure 3). Applying the Pythagorean theorem to this triangle we have that BQtwo + OBtwo = OQtwo. But it has already been said that BQ = Tan (α), that OB = 1 and that OQ = Sec (α), so that substituting in the Pythagorean equality for the right triangle OBQ we have:
Sotwo(α) + 1 = Sectwo(α).
Check whether or not the Pythagorean identities are fulfilled in the right triangle with legs AB = 4 and BC = 3.
Solution: The legs are known, the hypotenuse needs to be determined, which is:
AC = √ (AB ^ 2 + BC ^ 2) = √ (4 ^ 2 + 3 ^ 2) = √ (16 + 9) = √ (25) = 5.
The angle ∡BAC will be called α, ∡BAC = α. Now the trigonometric ratios are determined:
Sen α = BC / AC = 3/5
Cos α = AB / AC = 4/5
So α = BC / AB = 3/4
Cotan α = AB / BC = 4/3
Sec α = AC / AB = 5/4
Csc α = AC / BC = 5/3
It begins with the fundamental trigonometric identity:
Sentwo(α) + Costwo(α) = 1
(3/5) ^ 2 + (4/5) ^ 2 = 9/25 + 16/25 = (9 +16) / 25 = 25/25 = 1
It is concluded that it is fulfilled.
- The next Pythagorean identity is that of the tangent:
Sotwo(α) + 1 = Sectwo(α)
(3/4) ^ 2 + 1 = 9/16 + 16/16 = (9 + 16) / 16 = 25/16 = (5/4) ^ 2
And it is concluded that the identity of the tangent is verified.
- In a similar way that of the cotangent:
1 + Ctgtwo(α) = Csctwo(α)
1+ (4/3) ^ 2 = 1 + 16/9 = 25/9 = (5/3) ^ 2
It is concluded that it is also fulfilled, with which the task of verifying the Pythagorean identities for the given triangle has been completed.
Prove the following identities, based on the definitions of the trigonometric ratios and the Pythagorean identities.
Prove that Costwo x = (1 + Sen x) (1 - Sen x).
Solution: In the right side the notable product of the multiplication of a binomial by its conjugate is recognized, which, as is known, is a difference of squares:
Costwo x = 1two - Sentwo x
Then the term with sine on the right side passes to the left side with the sign changed:
Costwo x + Sentwo x = 1
Noting that the fundamental trigonometric identity has been reached, so it is concluded that the given expression is an identity, that is, it is true for any value of x.
Starting from the fundamental trigonometric identity and using the definitions of the trigonometric ratios, demonstrate the Pythagorean identity of the cosecant.
Solution: The fundamental identity is:
Sentwo(x) + Costwo(x) = 1
Both members are divided between Sentwo(x) and the denominator is distributed in the first member:
Sentwo(x) / Sentwo(x) + Costwo(x) / Sentwo(x) = 1 / Sentwo(x)
It is simplified:
1 + (Cos (x) / Sen (x)) ^ 2 = (1 / Sen (x)) ^ 2
Cos (x) / Sen (x) = Cotan (x) is a (non-Pythagorean) identity that is verified by the definition of the trigonometric ratios. The same happens with the following identity: 1 / Sen (x) = Csc (x).
Finally you have to:
1 + Ctgtwo(x) = Csctwo(x)
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