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Explain the Pythagorean identity in terms of the unit circle.

User UseRj
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The three Pythagorean trigonometric identities, which I’m sure one can find in any Algebra-Trigonometry textbook, are as follows:

sin² θ + cos² θ = 1

tan² θ + 1 = sec² θ

1 + cot² θ = csc² θ

where angle θ is any angle in standard position in the xy-plane.

Consistent with the definition of an identity, the above identities are true for all values of the variable, in this case angle θ, for which the functions involved are defined.

The Pythagorean Identities are so named because they are ultimately derived from a utilization of the Pythagorean Theorem, i.e., c² = a² + b², where c is the length of the hypotenuse of a right triangle and a and b are the lengths of the other two sides.

This derivation can be easily seen when considering the special case of the unit circle (r = 1). For any angle θ in standard position in the xy-plane and whose terminal side intersects the unit circle at the point (x, y), that is a distance r = 1 from the origin, we can construct a right triangle with hypotenuse c = r, with height a = y and with base b = x so that:

c² = a² + b² becomes:

r² = y² + x² = 1²

y² + x² = 1

We also know from our study of the unit circle that x = r(cos θ) = (1)(cos θ) = cos θ and y = r(sin θ) = (1)(sin θ) = sin θ; therefore, substituting, we get:

(sin θ)² + (cos θ)² = 1

1.) sin² θ + cos² θ = 1 which is the first Pythagorean Identity.

Now, if we divide through equation 1.) by cos² θ, we get the second Pythagorean Identity as follows:

(sin² θ + cos² θ)/cos² θ = 1/cos² θ

(sin² θ/cos² θ) + (cos² θ/cos² θ) = 1/cos² θ

(sin θ/cos θ)² + 1 = (1/cos θ)²

(tan θ)² + 1 = (sec θ)²

2.) tan² θ + 1 = sec² θ

Now, if we divide through equation 1.) by sin² θ, we get the third Pythagorean Identity as follows:

(sin² θ + cos² θ)/sin² θ = 1/sin² θ

(sin² θ/sin² θ) + (cos² θ/sin² θ) = 1/sin² θ

1 + (cos θ/sin θ)² = (1/sin θ)²

1 + (cot θ)² = (csc θ)²

3.) 1 + cot² θ = csc² θ

User Vantian
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