두 점 사이의 거리
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1. 개요
두 점 사이의 거리는 좌표 평면에서 두 점 사이의 길이를 계산하는 데 사용되는 수학적 개념이다. 삼각함수의 덧셈정리를 유도하는데 사용되며, 두 점 사이의 거리를 이용하여 삼각함수 계산, 삼각함수 방정식 및 부등식 풀이, 물리학 및 공학 문제 해결 등에 활용된다.
2. 삼각함수의 덧셈정리
300px l = \sqrt {({x_2}-{x_1})^2+({y_2}-{y_1})^2} 이므로,P= (cos \; \alpha, sin \; \alpha) \;\; , \;\; Q= (cos \beta, sin \beta) \overline{PQ}^2= (cos \beta - cos \; \alpha )^2 + ( sin \beta - sin \; \alpha)^2 = \left ( (cos \beta - cos \; \alpha ) \cdot (cos \beta - cos \; \alpha ) \right) + \left( ( sin \beta - sin \; \alpha) \cdot (- sin \beta - sin \; \alpha) \right) = \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) = (cos \beta)^2 + (cos \; \alpha)^2 + ( sin \beta)^2 + (sin \alpha)^2 - 2cos \alpha cos \beta -2 sin\alpha sin \beta = (cos^2 \beta + cos \; \alpha^2 ) + ( sin^2 \beta + sin^2 \alpha ) - 2 \left( cos \alpha cos \beta + sin\alpha sin \beta \right) \sin^2{x} + \cos^2{x} = 1 = 1 + 1 - 2 \left( cos \alpha cos \beta + sin\alpha sin \beta \right) \overline{PQ}^2= 2 - 2 \left( cos \alpha cos \beta + sin\alpha sin \beta \right) \overline{PQ}^2= \overline{OP}^2 +\overline{OQ}^2 - 2 \left(\overline{OP}\cdot {\overline{OQ} cos (\alpha-\beta)} \right) \overline{PQ}^2= 1^2 +1^2 - 2 \left(1\cdot {1 cos (\alpha-\beta)} \right) \overline{PQ}^2= 2 - 2 \left({ cos (\alpha-\beta)} \right) \overline{PQ}^2= 2 - 2 \left({ cos (\alpha-\beta)} \right)= 2 - 2 \left( cos \alpha cos \beta + sin\alpha sin \beta \right) cos \left({ \alpha-\beta} \right)= \left( cos \alpha cos \beta + sin\alpha sin \beta \right)
2. 1. 유도 과정
l = \sqrt {({x_2}-{x_1})^2+({y_2}-{y_1})^2} 이므로,P= (cos \; \alpha, sin \; \alpha) \;\; , \;\; Q= (cos \beta, sin \beta) \overline{PQ}^2= (cos \beta - cos \; \alpha )^2 + ( sin \beta - sin \; \alpha)^2 = \left ( (cos \beta - cos \; \alpha ) \cdot (cos \beta - cos \; \alpha ) \right) + \left( ( sin \beta - sin \; \alpha) \cdot (- sin \beta - sin \; \alpha) \right) = \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) = (cos \beta)^2 + (cos \; \alpha)^2 + ( sin \beta)^2 + (sin \alpha)^2 - 2cos \alpha cos \beta -2 sin\alpha sin \beta = (cos^2 \beta + cos \; \alpha^2 ) + ( sin^2 \beta + sin^2 \alpha ) - 2 \left( cos \alpha cos \beta + sin\alpha sin \beta \right) \sin^2{x} + \cos^2{x} = 1 = 1 + 1 - 2 \left( cos \alpha cos \beta + sin\alpha sin \beta \right) \overline{PQ}^2= 2 - 2 \left( cos \alpha cos \beta + sin\alpha sin \beta \right) \overline{PQ}^2= \overline{OP}^2 +\overline{OQ}^2 - 2 \left(\overline{OP}\cdot {\overline{OQ} cos (\alpha-\beta)} \right) \overline{PQ}^2= 1^2 +1^2 - 2 \left(1\cdot {1 cos (\alpha-\beta)} \right) \overline{PQ}^2= 2 - 2 \left({ cos (\alpha-\beta)} \right) \overline{PQ}^2= 2 - 2 \left({ cos (\alpha-\beta)} \right)= 2 - 2 \left( cos \alpha cos \beta + sin\alpha sin \beta \right) cos \left({ \alpha-\beta} \right)= \left( cos \alpha cos \beta + sin\alpha sin \beta \right) 2. 1. 1. 두 점 사이의 거리를 이용한 유도
l = \sqrt {({x_2}-{x_1})^2+({y_2}-{y_1})^2} 이므로,P= (cos \; \alpha, sin \; \alpha) \;\; , \;\; Q= (cos \beta, sin \beta) \overline{PQ}^2= (cos \beta - cos \; \alpha )^2 + ( sin \beta - sin \; \alpha)^2 = \left ( (cos \beta - cos \; \alpha ) \cdot (cos \beta - cos \; \alpha ) \right) + \left( ( sin \beta - sin \; \alpha) \cdot (- sin \beta - sin \; \alpha) \right) = \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) = (cos \beta)^2 + (cos \; \alpha)^2 + ( sin \beta)^2 + (sin \alpha)^2 - 2cos \alpha cos \beta -2 sin\alpha sin \beta = (cos^2 \beta + cos \; \alpha^2 ) + ( sin^2 \beta + sin^2 \alpha ) - 2 \left( cos \alpha cos \beta + sin\alpha sin \beta \right) \sin^2{x} + \cos^2{x} = 1 = 1 + 1 - 2 \left( cos \alpha cos \beta + sin\alpha sin \beta \right) \overline{PQ}^2= 2 - 2 \left( cos \alpha cos \beta + sin\alpha sin \beta \right) \overline{PQ}^2= \overline{OP}^2 +\overline{OQ}^2 - 2 \left(\overline{OP}\cdot {\overline{OQ} cos (\alpha-\beta)} \right) \overline{PQ}^2= 1^2 +1^2 - 2 \left(1\cdot {1 cos (\alpha-\beta)} \right) \overline{PQ}^2= 2 - 2 \left({ cos (\alpha-\beta)} \right) \overline{PQ}^2= 2 - 2 \left({ cos (\alpha-\beta)} \right)= 2 - 2 \left( cos \alpha cos \beta + sin\alpha sin \beta \right) cos \left({ \alpha-\beta} \right)= \left( cos \alpha cos \beta + sin\alpha sin \beta \right) 2. 1. 2. 제2 코사인 법칙을 이용한 유도
l = \sqrt {({x_2}-{x_1})^2+({y_2}-{y_1})^2} 이므로,P= (cos \; \alpha, sin \; \alpha) \;\; , \;\; Q= (cos \beta, sin \beta) \overline{PQ}^2= (cos \beta - cos \; \alpha )^2 + ( sin \beta - sin \; \alpha)^2 = \left ( (cos \beta - cos \; \alpha ) \cdot (cos \beta - cos \; \alpha ) \right) + \left( ( sin \beta - sin \; \alpha) \cdot (- sin \beta - sin \; \alpha) \right) = \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) = (cos \beta)^2 + (cos \; \alpha)^2 + ( sin \beta)^2 + (sin \alpha)^2 - 2cos \alpha cos \beta -2 sin\alpha sin \beta = (cos^2 \beta + cos \; \alpha^2 ) + ( sin^2 \beta + sin^2 \alpha ) - 2 \left( cos \alpha cos \beta + sin\alpha sin \beta \right) \sin^2{x} + \cos^2{x} = 1 = 1 + 1 - 2 \left( cos \alpha cos \beta + sin\alpha sin \beta \right) \overline{PQ}^2= 2 - 2 \left( cos \alpha cos \beta + sin\alpha sin \beta \right) \overline{PQ}^2= \overline{OP}^2 +\overline{OQ}^2 - 2 \left(\overline{OP}\cdot {\overline{OQ} cos (\alpha-\beta)} \right) \overline{PQ}^2= 1^2 +1^2 - 2 \left(1\cdot {1 cos (\alpha-\beta)} \right) \overline{PQ}^2= 2 - 2 \left({ cos (\alpha-\beta)} \right) \overline{PQ}^2= 2 - 2 \left({ cos (\alpha-\beta)} \right)= 2 - 2 \left( cos \alpha cos \beta + sin\alpha sin \beta \right) cos \left({ \alpha-\beta} \right)= \left( cos \alpha cos \beta + sin\alpha sin \beta \right) 2. 1. 3. 기하학적 증명
l = \sqrt {({x_2}-{x_1})^2+({y_2}-{y_1})^2} 이므로,P= (cos \; \alpha, sin \; \alpha) \;\; , \;\; Q= (cos \beta, sin \beta) \overline{PQ}^2= (cos \beta - cos \; \alpha )^2 + ( sin \beta - sin \; \alpha)^2 = \left ( (cos \beta - cos \; \alpha ) \cdot (cos \beta - cos \; \alpha ) \right) + \left( ( sin \beta - sin \; \alpha) \cdot (- sin \beta - sin \; \alpha) \right) = \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) = (cos \beta)^2 + (cos \; \alpha)^2 + ( sin \beta)^2 + (sin \alpha)^2 - 2cos \alpha cos \beta -2 sin\alpha sin \beta = (cos^2 \beta + cos \; \alpha^2 ) + ( sin^2 \beta + sin^2 \alpha ) - 2 \left( cos \alpha cos \beta + sin\alpha sin \beta \right) \sin^2{x} + \cos^2{x} = 1 = 1 + 1 - 2 \left( cos \alpha cos \beta + sin\alpha sin \beta \right) \overline{PQ}^2= 2 - 2 \left( cos \alpha cos \beta + sin\alpha sin \beta \right) \overline{PQ}^2= \overline{OP}^2 +\overline{OQ}^2 - 2 \left(\overline{OP}\cdot {\overline{OQ} cos (\alpha-\beta)} \right) \overline{PQ}^2= 1^2 +1^2 - 2 \left(1\cdot {1 cos (\alpha-\beta)} \right) \overline{PQ}^2= 2 - 2 \left({ cos (\alpha-\beta)} \right) \overline{PQ}^2= 2 - 2 \left({ cos (\alpha-\beta)} \right)= 2 - 2 \left( cos \alpha cos \beta + sin\alpha sin \beta \right) cos \left({ \alpha-\beta} \right)= \left( cos \alpha cos \beta + sin\alpha sin \beta \right) 2. 2. 공식 목록
3. 덧셈정리의 활용
3. 1. 삼각함수 계산
3. 2. 삼각함수 방정식 및 부등식
3. 3. 물리학에서의 활용
3. 4. 공학에서의 활용
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