두 점 사이의 거리: 두 판 사이의 차이
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19번째 줄:
:<math>l = \sqrt {({x_2}-{x_1})^2+({y_2}-{y_1})^2}</math>
==삼각함수의 덧셈정리==
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[[File:DistanceFromAtoB002.svg|300px]]
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:<math>l = \sqrt {({x_2}-{x_1})^2+({y_2}-{y_1})^2}</math>이므로,
:<math>P= (cos \; \alpha, sin \; \alpha) \;\; , \;\; Q= (cos \beta, sin \beta)</math>
:<math>\overline{PQ}^2= (cos \beta - cos \; \alpha )^2 + ( sin \beta - sin \; \alpha)^2 </math>
:<math>= \left ( (cos \beta - cos \; \alpha ) \cdot (cos \beta - cos \; \alpha ) \right) + \left( ( sin \beta - sin \; \alpha) \cdot (- sin \beta - sin \; \alpha) \right) </math>
:<math>= \left ( (cos \beta)^2 - 2cos \alpha cos \beta + (cos \; \alpha)^2 \right) + \left( ( sin \beta)^2 -2 sin\alpha sin \beta + (sin \alpha)^2 \right) </math>
:<math>= (cos \beta)^2 + (cos \; \alpha)^2 + ( sin \beta)^2 + (sin \alpha)^2 - 2cos \alpha cos \beta -2 sin\alpha sin \beta </math>
:<math>= (cos^2 \beta + cos \; \alpha^2 ) + ( sin^2 \beta + sin^2 \alpha ) - 2 \left( cos \alpha cos \beta + sin\alpha sin \beta \right)</math>
그리고 [[삼각함수 항등식]]의 [[피타고라스 정리]]에서,
: <math> \sin^2{x} + \cos^2{x} = 1 </math>
따라서,
:<math>= 1 + 1 - 2 \left( cos \alpha cos \beta + sin\alpha sin \beta \right)</math>
:<math>\overline{PQ}^2= 2 - 2 \left( cos \alpha cos \beta + sin\alpha sin \beta \right)</math>
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한편,
:이것은,[[코사인법칙#제2코사인법칙|제2코사인법칙]]에서는,
:<math>\overline{PQ}^2= \overline{OP}^2 +\overline{OQ}^2 - 2 \left(\overline{OP}\cdot {\overline{OQ} cos (\alpha-\beta} \right) </math>
:<math>\overline{PQ}^2= 1^2 +1^2 - 2 \left(1\cdot {1 cos (\alpha-\beta} \right) </math>
:<math>\overline{PQ}^2= 2 - 2 \left({ cos \alpha-\beta} \right) </math>
그리고,
:<math>\overline{PQ}^2= 2 - 2 \left({ cos \alpha-\beta} \right)= 2 - 2 \left( cos \alpha cos \beta + sin\alpha sin \beta \right) </math>
따라서,
:<math> \left({ cos \alpha-\beta} \right)= \left( cos \alpha cos \beta + sin\alpha sin \beta \right) </math>
==함께보기==
*[[거리공간]]
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