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受欢迎的三角函数 >

14.79m=((7.61_{0}^2sin(2θ_{0}))/g)

解答

14.79 m = (\frac{7.6 1 _{0}^{2} sin ( 2 θ _{0} )}{g})

解答

θ_{0} = \frac{arcsin ( 0.25538 \dots m g )}{2} + πn, θ_{0} = \frac{π}{2} + \frac{arcsin ( - 0.25538 \dots m g )}{2} + πn

求解步骤

14.79 m = (\frac{7.6 1 _{0}^{2} sin ( 2 θ _{0} )}{g})

交换两边

\frac{7.6 1 _{0}^{2} sin ( 2 θ _{0} )}{g} = 14.79 m

在两边乘以

g

\frac{7.6 1 _{0}^{2} sin ( 2 θ _{0} )}{g} = 14.79 m

在两边乘以

g

\frac{7.6 1 _{0}^{2} sin ( 2 θ _{0} ) g}{g} = 14.79 m g; g \neq = 0

化简

57.9121 sin (2 θ_{0}) = 14.79 m g; g \neq = 0

57.9121 sin (2 θ_{0}) = 14.79 m g; g \neq = 0

两边除以

57.9121; g \neq = 0

57.9121 sin (2 θ_{0}) = 14.79 m g; g \neq = 0

两边除以

57.9121; g \neq = 0

\frac{57.9121 sin ( 2 θ _{0} )}{57.9121} = \frac{14.79 m g}{57.9121}; g \neq = 0

化简

sin (2 θ_{0}) = 0.25538 \dots m g; g \neq = 0

sin (2 θ_{0}) = 0.25538 \dots m g; g \neq = 0

使用反三角函数性质

sin (2 θ_{0}) = 0.25538 \dots m g

sin (2 θ_{0}) = 0.25538 \dots m g

的通解

sin (x) = a \Rightarrow x = arcsin (a) + 2 πn, x = π + arcsin (a) + 2 πn

2 θ_{0} = arcsin (0.25538 \dots m g) + 2 πn, 2 θ_{0} = π + arcsin (- 0.25538 \dots m g) + 2 πn

2 θ_{0} = arcsin (0.25538 \dots m g) + 2 πn, 2 θ_{0} = π + arcsin (- 0.25538 \dots m g) + 2 πn

解

2 θ_{0} = arcsin (0.25538 \dots m g) + 2 πn : θ_{0} = \frac{arcsin ( 0.25538 \dots m g )}{2} + πn

2 θ_{0} = arcsin (0.25538 \dots m g) + 2 πn

两边除以

2

2 θ_{0} = arcsin (0.25538 \dots m g) + 2 πn

两边除以

2

\frac{2 θ _{0}}{2} = \frac{arcsin ( 0.25538 \dots m g )}{2} + \frac{2 πn}{2}

化简

θ_{0} = \frac{arcsin ( 0.25538 \dots m g )}{2} + πn

θ_{0} = \frac{arcsin ( 0.25538 \dots m g )}{2} + πn

解

2 θ_{0} = π + arcsin (- 0.25538 \dots m g) + 2 πn : θ_{0} = \frac{π}{2} + \frac{arcsin ( - 0.25538 \dots m g )}{2} + πn

2 θ_{0} = π + arcsin (- 0.25538 \dots m g) + 2 πn

两边除以

2

2 θ_{0} = π + arcsin (- 0.25538 \dots m g) + 2 πn

两边除以

2

\frac{2 θ _{0}}{2} = \frac{π}{2} + \frac{arcsin ( - 0.25538 \dots m g )}{2} + \frac{2 πn}{2}

化简

θ_{0} = \frac{π}{2} + \frac{arcsin ( - 0.25538 \dots m g )}{2} + πn

θ_{0} = \frac{π}{2} + \frac{arcsin ( - 0.25538 \dots m g )}{2} + πn

θ_{0} = \frac{arcsin ( 0.25538 \dots m g )}{2} + πn, θ_{0} = \frac{π}{2} + \frac{arcsin ( - 0.25538 \dots m g )}{2} + πn

作图

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