Derivation of the Pythagorean Trigonometric Identities

Pronunciation: /ˌdɛɹ.ɪˈveɪ.ʃən ʌv ðə pɪˌθæ.gəˈri.ən ˌtrɪɡ.ə.nəˈmɛ.trɪk aɪˈdɛn.tə.tiz/ Explain

Derivation of sin^2 theta + cos^2 theta = 1
ImageEquationDiscussion
A2 + B2 = C2 This is the equation given by the Pythagorean Theorem.
(A^2 + B^2)/C^2 = C^2/C^2 Use the multiplicative property of equality to multiply both sides by (A^2 + B^2)/C^2 = 1.
(A^2 + B^2)/C^2 = 1 Since anything except zero divided by itself is one, substitute 1 for (C^2)/(C^2).
(C^2)/(C^2) Use the distributive property of multiplication over addition and subtraction to distribute (C^2)/(C^2) through the parentheses on the left side of the equation.
(A/C)^2 + (B/C)^2 = 1 Apply the distributive property of exponentiation over division.
sin^2 theta + cos^2 theta = 1 The definition of sine is sine theta = opposite/hypotenuse. The definition of cosine is cosine theta = adjacent/hypotenuse. Given angle θ, sin theta = A/C and cos theta = B/C. Apply the substitution property of equality to substitute sin θ for A/C and cos θ for B/C.

Derivation of tan^2 theta + 1 = sec^2 theta
ImageEquationDiscussion
sin^2 theta + cos^2 theta = 1 Start with the first Pythagorean identity.
(sin^2 theta + cos^2 theta)/(cos^2 theta) = 1/(cos^2 theta) Apply the multiplication property of equality to the equation by multiplying both sides of the equation by (sin^2 theta)/(cos^2 theta) + (cos^2 theta)/(cos^2 theta) = 1/(cos^2 theta).
(sin^2 theta)/(cos^2 theta) + (cos^2 theta)/(cos^2 theta) = 1/(cos^2 theta) Apply the distributive property of multiplication over addition and subtraction.
(sin^2 theta)/(cos^2 theta) + 1 = 1/(cos^2 theta) Since anything except zero divided by itself is one, substitute 1 for (sin^2 theta)/(cos^2 theta) + 1 = (1^2)/(cos^2 theta).
(sin^2 theta)/(cos^2 theta) + 1 = (1^2)/(cos^2 theta) Since 12 = 1, apply the substiution property of equality to subtitute 12 = 1 for 1.
((sin theta)/(cos theta))^2 + 1 = (1/(cos theta))^2 Apply the distributive property of exponentiation over division to both sides of the equation.
tan^2 theta + 1 = (1/(cos theta))^2 Use the definition of tangent to subsitute tan θ for (sin theta)/(cos theta).
tan^2 theta + 1 = sec^2 theta Use the definition of secant to subsitute sec θ for 1/(cos theta).

Derivation of (sin^2 theta + cos^2 theta)/(sin^2 theta) = 1/(sin^2 theta)
ImageEquationDiscussion
sin^2 theta + cos^2 theta = 1 Start with the first Pythagorean identity.
(sin^2 theta + cos^2 theta)/(sin^2 theta) = 1/(sin^2 theta) Apply the multiplication property of equality to the equation by multiplying both sides of the equation by (sin^2 theta)/(cos^2 theta) + (cos^2 theta)/(cos^2 theta) = 1/(cos^2 theta).
(sin^2 theta)/(sin^2 theta) + (cos^2 theta)/(sin^2 theta) = 1/(sin^2 theta) Apply the distributive property of multiplication over addition and subtraction.
1 + (cos^2 theta)/(sin^2 theta) = 1/(sin^2 theta) Since anything except zero divided by itself is one, substitute 1 for (sin^2 theta)/(sin^2 theta).
(cos^2 theta)/(sin^2 theta) + 1 = 1/(sin^2 theta) Apply the commutative property of addition.
(cos^2 theta)/(sin^2 theta) + 1 = (1^2)/(sin^2 theta) Since 12 = 1, apply the substiution property of equality to subtitute 12 = 1 for 1.
((cos theta)/(sin theta))^2 + 1 = (1/(sin theta))^2 Apply the distributive property of exponentiation over division to both sides of the equation.
cot^2 theta + 1 = (1/(cos theta))^2 Use the definition of tangent to subsitute cot θ for (cos theta)/(sin theta).
cot^2 theta + 1 = csc^2 theta Use the definition of secant to subsitute csc θ for 1/(sin theta).

References

  1. McAdams, David E.. All Math Words Dictionary, derivation. 2nd Classroom edition 20150108-4799968. pg 58. Life is a Story Problem LLC. January 8, 2015. Buy the book

Cite this article as:

 McAdams, David E. Derivation of the Pythagorean Trigonometric Identities. 12/21/2018. All Math Words Encyclopedia. Life is a Story Problem LLC. https://www.allmathwords.org/en/t/ti_pythagorean.html.

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Revision History

12/21/2018: Reviewed and corrected IPA pronunication. (McAdams, David E.)
7/4/2018: Removed broken links, updated license, implemented new markup, implemented new Geogebra protocol. (McAdams, David E.)
4/29/2011: Initial version. (McAdams, David E.)

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