What is the pythagorean identity?

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Answer 1 · 2014-11-02T12:01:40

Pythagorean Identity

${cos}^{2} θ + {sin}^{2} θ = 1$ $cos^2theta+sin^2theta=1$

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Answer 2 · 2016-01-08T21:37:17

The Pythagorean identity is:

${sin}^{2} x + {cos}^{2} x = 1$ $color(red)(sin^2x+cos^2x=1$

However, it does not have to apply to just sine and cosine.

To find the form of the Pythagorean identity with the other trigonometric identities, divide the original identity by sine and cosine.

SINE:

$\frac{{sin}^{2} x + {cos}^{2} x = 1}{{sin}^{2} x}$ $(sin^2x+cos^2x=1)/sin^2x$

This gives:

$\frac{{sin}^{2} x}{{sin}^{2} x} + \frac{{cos}^{2} x}{{sin}^{2} x} = \frac{1}{{sin}^{2} x}$ $sin^2x/sin^2x+cos^2x/sin^2x=1/sin^2x$

Which equals

$1 + {cot}^{2} x = {csc}^{2} x$ $color(red)(1+cot^2x=csc^2x$

To find the other identity:

COSINE:

$\frac{{sin}^{2} x + {cos}^{2} x = 1}{{cos}^{2} x}$ $(sin^2x+cos^2x=1)/cos^2x$

This gives:

$\frac{{sin}^{2} x}{{cos}^{2} x} + \frac{{cos}^{2} x}{{cos}^{2} x} = \frac{1}{{cos}^{2} x}$ $sin^2x/cos^2x+cos^2x/cos^2x=1/cos^2x$

Which equals

${tan}^{2} x + 1 = {sec}^{2} x$ $color(red)(tan^2x+1=sec^2x$

These identities can all be algebraically manipulated to prove many things:

${(sin^2x=1-cos^2x),(cos^2x=1-sin^2x):}$

${(tan^2x=sec^2x-1),(cot^2x=csc^2x-1):}$