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Difference between revisions of "Pythagorean identities"
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<math>\tan^2x + 1 = \sec^2x</math>
 
<math>\tan^2x + 1 = \sec^2x</math>
  
−
Using the unit circle definition of trigonometry, because the point <math>(\cos (x), \sin (x))</math> is defined to be on the unit circle, it is a distance one away from the origin. Then by the distance formula, <math>\sin^2x + \cos^2x = 1</math>. We can also think of it the following way. Suppose that there is a right triangle <math>ABC</math> with the right angle at <math>B</math>. Then, we have:
+
Using the unit circle definition of trigonometry, because the point <math>(\cos (x), \sin (x))</math> is defined to be on the unit circle, it is a distance one away from the origin. Then by the distance formula, <math>\sin^2x + \cos^2x = 1</math>.
 +
 
 +
Another way to think of it is as follows: Suppose that there is a right triangle <math>ABC</math> with the right angle at <math>B</math>. Then, we have:
  
 
<math>BC^2+AB^2=AC^2</math>
 
<math>BC^2+AB^2=AC^2</math>
Line 14: Line 16:
  
 
<math>\sin^2A+cos^2A+1</math>.
 
<math>\sin^2A+cos^2A+1</math>.
−
To derive the other two Pythagorean identities, divide by either <math>\sin^2 (x)</math> or <math>\cos^2 (x)</math> and substitute the respective trigonometry in place of the ratios to obtain the desired result.
+
 
 +
To derive the other two Pythagorean identities, divide <math>\sin^2A+cos^2A+1</math> by either <math>\sin^2 (x)</math> or <math>\cos^2 (x)</math> and substitute the respective trigonometry in place of the ratios to obtain the desired result.
  
 
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Revision as of 13:06, 3 January 2024

The Pythagorean identities state that

$\sin^2x + \cos^2x = 1$

$1 + \cot^2x = \csc^2x$

$\tan^2x + 1 = \sec^2x$

Using the unit circle definition of trigonometry, because the point $(\cos (x), \sin (x))$ is defined to be on the unit circle, it is a distance one away from the origin. Then by the distance formula, $\sin^2x + \cos^2x = 1$.

Another way to think of it is as follows: Suppose that there is a right triangle $ABC$ with the right angle at $B$. Then, we have:

$BC^2+AB^2=AC^2$

$\frac{BC^2}{AC^2}+\frac{AB^2}{AC^2}+\frac{AC^2}{AC^2}$

$\sin^2A+cos^2A+1$.

To derive the other two Pythagorean identities, divide $\sin^2A+cos^2A+1$ by either $\sin^2 (x)$ or $\cos^2 (x)$ and substitute the respective trigonometry in place of the ratios to obtain the desired result.

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See Also

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