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2.例题例1.己知扇形的弧含有54,半径等于20厘米;求扇形的局长和面积。(精确到1) 解:设弧长为L,半径为R、含中心角a弧度, 则a/54°=π/180°=3π/10,因此 L=aR=(3π/10)×20=6π。故周长≈59(厘米)面积=(1/2)RL≈190(平方厘米) 例2。设a是第四象限的角,比较sina和tga的值的大小。解:如图1,角a的终边落在第四象限,与单位圆交于P,与过n的切线交于T,作PM⊥OA于M,则sina=MP,tga=AT。因为|MP|<|AT|,∴MP>AT,故sina>tga。例3.已知cosa=m,求sina和tga的值。解:当a是第一、二象限的角或a=kπ(k∈Z)时、sina=(1-m~2)~(1/2),tga=(1-m~2)~(1/2)/m
2. Examples Example 1. The arc of the known fan contains 54 and the radius is equal to 20 cm; (accurate to 1) Solution: Let arc length L, radius R, and center angle a radians, then a/54°=π/180°=3π/10, so L=aR=(3π/10)×20 =6π. Therefore, perimeter 59 (cm) area = (1/2) RL ≈ 190 (cm 2) Example 2. Let a be the angle of the fourth quadrant, compare the size of the sina and tga values. Solution: As shown in Figure 1, the end of the angle a falls in the fourth quadrant, intersects with the unit circle at P, and intersects with the tangent of n over T, for PM⊥OA in M, then sina=MP,tga=AT. Because |MP|<|AT|, ∴MP>AT, sina>tga. Example 3. Knowing cosa=m, find the values of sina and tga. Solution: When a is the angle of the first and second quadrants or a=kπ(k∈Z), sina=(1-m~2)~(1/2), tga=(1-m~2)~( 1/2)/m