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The Pythagorean trigonometric identity says that for any angle A:
- <math>\sin^2 A + \cos^2 A = 1<math>
Proof
- <math>c=\sqrt{a^2+b^2}<math>
- <math>\begin{matrix}\sin A & = & \frac{a}{c} \\ \ & = & \frac{a}{\sqrt{a^2+b^2}} \end{matrix}<math>
- <math>\begin{matrix}\cos A & = & \frac{b}{c} \\ \ & = & \frac{b}{\sqrt{a^2+b^2}} \end{matrix}<math>
- <math>\begin{matrix}\sin^2(A) + \cos^2(A) & = & \left( \frac{a}{c} \right)^2 + \left( \frac{b}{c} \right)^2 \\ \ & = & \left ( \frac{a}{\sqrt{a^2 + b^2}} \right )^2 + \left ( \frac{b}{\sqrt{a^2 + b^2}} \right )^2 \\ \ & = & \left ( \frac{a}{\sqrt{a^2 + b^2}} \times \frac{a}{\sqrt{a^2 + b^2}} \right ) + \left ( \frac{b}{\sqrt{a^2 + b^2}} \times \frac{b}{\sqrt{a^2 + b^2}} \right ) \\ \ & = & \frac{a^2}{\left(\sqrt{a^2 + b^2}\right)^2} + \frac{b^2}{\left(\sqrt{a^2 + b^2}\right)^2} \\ \ & = & \frac{a^2}{a^2 + b^2} + \frac{b^2}{a^2 + b^2} \\ \ & = & \frac{a^2 + b^2}{a^2 + b^2} \\ \ & = & 1\end{matrix}<math>
Or:
- <math>\begin{matrix}\sin^2(A) + \cos^2(A) & = & \left( \frac{a}{c} \right)^2 + \left( \frac{b}{c} \right)^2 \\ \ & = & \left ( \frac{a}{c} \times \frac{a}{c} \right ) + \left ( \frac{b}{c} \times \frac{b}{c} \right ) \\ \ & = & \frac{a^2}{c^2} + \frac{b^2}{c^2} \\ \ & = & \frac{a^2 + b^2}{c^2} \\ \ & = & \frac{a^2 + b^2}{\left(\sqrt{a^2 + b^2}\right)^2} \\ \ & = & \frac{a^2 + b^2}{a^2 + b^2} \\ \ & = & 1\end{matrix}<math>
Note
The reason for:
- <math>\frac{a^2 + b^2}{a^2 + b^2} = 1<math>
is that any number, when divided by itself, is equal to one.
See also
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