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受欢迎的三角函数 >

solvefor θ,r=vcos(θ)((2vsin(θ))/g)

解答

求解

θ, r = v cos (θ) (\frac{2 v sin ( θ )}{g})

解答

θ = \frac{arcsin ( \frac{g r}{v ^{2}} )}{2} + πn, θ = \frac{π}{2} + \frac{arcsin ( - \frac{g r}{v ^{2}} )}{2} + πn

求解步骤

r = v cos (θ) (\frac{2 v sin ( θ )}{g})

交换两边

v cos (θ) \frac{2 v sin ( θ )}{g} = r

两边减去

r

\frac{2 v ^{2} sin ( θ ) cos ( θ )}{g} - r = 0

化简

\frac{2 v ^{2} sin ( θ ) cos ( θ )}{g} - r : \frac{2 v ^{2} sin ( θ ) cos ( θ ) - g r}{g}

\frac{2 v ^{2} sin ( θ ) cos ( θ )}{g} - r

将项转换为分式:

r = \frac{r g}{g}

= \frac{2 v ^{2} sin ( θ ) cos ( θ )}{g} - \frac{r g}{g}

因为分母相等，所以合并分式:

\frac{a}{c} \pm \frac{b}{c} = \frac{a \pm b}{c}

= \frac{2 v ^{2} sin ( θ ) cos ( θ ) - r g}{g}

\frac{2 v ^{2} sin ( θ ) cos ( θ ) - g r}{g} = 0

\frac{f ( x )}{g ( x )} = 0 \Rightarrow f (x) = 0

2 v^{2} sin (θ) cos (θ) - g r = 0

使用三角恒等式改写

2 v^{2} sin (θ) cos (θ) - g r

使用倍角公式:

2 sin (x) cos (x) = sin (2 x)

= - g r + sin (2 θ) v^{2}

- g r + sin (2 θ) v^{2} = 0

将

g r

到右边

- g r + sin (2 θ) v^{2} = 0

两边加上

g r

- g r + sin (2 θ) v^{2} + g r = 0 + g r

化简

sin (2 θ) v^{2} = g r

sin (2 θ) v^{2} = g r

两边除以

v^{2}; v \neq = 0

sin (2 θ) v^{2} = g r

两边除以

v^{2}; v \neq = 0

\frac{sin ( 2 θ ) v ^{2}}{v ^{2}} = \frac{g r}{v ^{2}}; v \neq = 0

化简

sin (2 θ) = \frac{g r}{v ^{2}}; v \neq = 0

sin (2 θ) = \frac{g r}{v ^{2}}; v \neq = 0

使用反三角函数性质

sin (2 θ) = \frac{g r}{v ^{2}}

sin (2 θ) = \frac{g r}{v ^{2}}

的通解

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

2 θ = arcsin (\frac{g r}{v ^{2}}) + 2 πn, 2 θ = π + arcsin (- \frac{g r}{v ^{2}}) + 2 πn

2 θ = arcsin (\frac{g r}{v ^{2}}) + 2 πn, 2 θ = π + arcsin (- \frac{g r}{v ^{2}}) + 2 πn

解

2 θ = arcsin (\frac{g r}{v ^{2}}) + 2 πn : θ = \frac{arcsin ( \frac{g r}{v ^{2}} )}{2} + πn

2 θ = arcsin (\frac{g r}{v ^{2}}) + 2 πn

两边除以

2

2 θ = arcsin (\frac{g r}{v ^{2}}) + 2 πn

两边除以

2

\frac{2 θ}{2} = \frac{arcsin ( \frac{g r}{v ^{2}} )}{2} + \frac{2 πn}{2}

化简

θ = \frac{arcsin ( \frac{g r}{v ^{2}} )}{2} + πn

θ = \frac{arcsin ( \frac{g r}{v ^{2}} )}{2} + πn

解

2 θ = π + arcsin (- \frac{g r}{v ^{2}}) + 2 πn : θ = \frac{π}{2} + \frac{arcsin ( - \frac{g r}{v ^{2}} )}{2} + πn

2 θ = π + arcsin (- \frac{g r}{v ^{2}}) + 2 πn

两边除以

2

2 θ = π + arcsin (- \frac{g r}{v ^{2}}) + 2 πn

两边除以

2

\frac{2 θ}{2} = \frac{π}{2} + \frac{arcsin ( - \frac{g r}{v ^{2}} )}{2} + \frac{2 πn}{2}

化简

θ = \frac{π}{2} + \frac{arcsin ( - \frac{g r}{v ^{2}} )}{2} + πn

θ = \frac{π}{2} + \frac{arcsin ( - \frac{g r}{v ^{2}} )}{2} + πn

θ = \frac{arcsin ( \frac{g r}{v ^{2}} )}{2} + πn, θ = \frac{π}{2} + \frac{arcsin ( - \frac{g r}{v ^{2}} )}{2} + πn

作图

流行的例子

1+tan^2(x)cos(x)=sec(x)

sec(θ)=(sqrt(6))/2

1.5sqrt(2)=3cos(x)