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受欢迎的三角函数 >

solvefor x,z=arcsin((xy)/(x^2+y^2))

解答

求解

x, z = arcsin (\frac{x y}{x ^{2} + y ^{2}})

解答

x = \frac{y + y 1 - 4 sin ^{2} ( z )}{2 sin ( z )}, x = \frac{y - y 1 - 4 sin ^{2} ( z )}{2 sin ( z )}

求解步骤

z = arcsin (\frac{x y}{x ^{2} + y ^{2}})

交换两边

arcsin (\frac{x y}{x ^{2} + y ^{2}}) = z

使用反三角函数性质

arcsin (\frac{x y}{x ^{2} + y ^{2}}) = z

arcsin (x) = a \Rightarrow x = sin (a)

\frac{x y}{x ^{2} + y ^{2}} = sin (z)

\frac{x y}{x ^{2} + y ^{2}} = sin (z)

解

\frac{x y}{x ^{2} + y ^{2}} = sin (z) : x = \frac{y + y 1 - 4 sin ^{2} ( z )}{2 sin ( z )}, x = \frac{y - y 1 - 4 sin ^{2} ( z )}{2 sin ( z )}; sin (z) \neq = 0

\frac{x y}{x ^{2} + y ^{2}} = sin (z)

在两边乘以

x^{2} + y^{2}

\frac{x y}{x ^{2} + y ^{2}} = sin (z)

在两边乘以

x^{2} + y^{2}

\frac{x y}{x ^{2} + y ^{2}} (x^{2} + y^{2}) = sin (z) (x^{2} + y^{2})

化简

x y = sin (z) (x^{2} + y^{2})

x y = sin (z) (x^{2} + y^{2})

解

x y = sin (z) (x^{2} + y^{2}) : x = \frac{y + y 1 - 4 sin ^{2} ( z )}{2 sin ( z )}, x = \frac{y - y 1 - 4 sin ^{2} ( z )}{2 sin ( z )}; sin (z) \neq = 0

x y = sin (z) (x^{2} + y^{2})

展开

sin (z) (x^{2} + y^{2}) : x^{2} sin (z) + y^{2} sin (z)

sin (z) (x^{2} + y^{2})

使用分配律:

a (b + c) = ab + a c

a = sin (z), b = x^{2}, c = y^{2}

= sin (z) x^{2} + sin (z) y^{2}

= x^{2} sin (z) + y^{2} sin (z)

x y = x^{2} sin (z) + y^{2} sin (z)

交换两边

x^{2} sin (z) + y^{2} sin (z) = x y

将

x y

para o lado esquerdo

x^{2} sin (z) + y^{2} sin (z) = x y

两边减去

x y

x^{2} sin (z) + y^{2} sin (z) - x y = x y - x y

化简

x^{2} sin (z) + y^{2} sin (z) - x y = 0

x^{2} sin (z) + y^{2} sin (z) - x y = 0

改写成标准形式

a x^{2} + b x + c = 0

sin (z) x^{2} - y x + y^{2} sin (z) = 0

使用求根公式求解

sin (z) x^{2} - y x + y^{2} sin (z) = 0

二次方程求根公式:

若

a = sin (z), b = - y, c = y^{2} sin (z)

x_{1, 2} = \frac{- ( - y ) \pm ( - y ) ^{2} - 4 sin ( z ) y ^{2} sin ( z )}{2 sin ( z )}

x_{1, 2} = \frac{- ( - y ) \pm ( - y ) ^{2} - 4 sin ( z ) y ^{2} sin ( z )}{2 sin ( z )}

化简

(- y)^{2} - 4 sin (z) y^{2} sin (z) : y 1 - 4 sin^{2} (z)

(- y)^{2} - 4 sin (z) y^{2} sin (z)

(- y)^{2} = y^{2}

(- y)^{2}

使用指数法则:

(- a)^{n} = a^{n},

若

n

是偶数

(- y)^{2} = y^{2}

= y^{2}

4 sin (z) y^{2} sin (z) = 4 y^{2} sin^{2} (z)

4 sin (z) y^{2} sin (z)

使用指数法则:

a^{b} \cdot a^{c} = a^{b + c}

sin (z) sin (z) = sin^{1 + 1} (z)

= 4 y^{2} sin^{1 + 1} (z)

数字相加:

1 + 1 = 2

= 4 y^{2} sin^{2} (z)

= y^{2} - 4 y^{2} sin^{2} (z)

因式分解出通项

y^{2}

= y^{2} (1 - 4 sin^{2} (z))

使用根式运算法则:

n ab = n a n b,

假定

a \geq 0, b \geq 0

= y^{2} - 4 sin^{2} (z) + 1

使用根式运算法则:

n a^{n} = a,

假定

a \geq 0

y^{2} = y

= y - 4 sin^{2} (z) + 1

x_{1, 2} = \frac{- ( - y ) \pm y 1 - 4 sin ^{2} ( z )}{2 sin ( z )}; sin (z) \neq = 0

将解分隔开

x_{1} = \frac{- ( - y ) + y 1 - 4 sin ^{2} ( z )}{2 sin ( z )}, x_{2} = \frac{- ( - y ) - y 1 - 4 sin ^{2} ( z )}{2 sin ( z )}

x = \frac{- ( - y ) + y 1 - 4 sin ^{2} ( z )}{2 sin ( z )} : \frac{y + y 1 - 4 sin ^{2} ( z )}{2 sin ( z )}

\frac{- ( - y ) + y 1 - 4 sin ^{2} ( z )}{2 sin ( z )}

使用法则

- (- a) = a

= \frac{y + y 1 - 4 sin ^{2} ( z )}{2 sin ( z )}

x = \frac{- ( - y ) - y 1 - 4 sin ^{2} ( z )}{2 sin ( z )} : \frac{y - y 1 - 4 sin ^{2} ( z )}{2 sin ( z )}

\frac{- ( - y ) - y 1 - 4 sin ^{2} ( z )}{2 sin ( z )}

使用法则

- (- a) = a

= \frac{y - y 1 - 4 sin ^{2} ( z )}{2 sin ( z )}

二次方程组的解是:

x = \frac{y + y 1 - 4 sin ^{2} ( z )}{2 sin ( z )}, x = \frac{y - y 1 - 4 sin ^{2} ( z )}{2 sin ( z )}; sin (z) \neq = 0

x = \frac{y + y 1 - 4 sin ^{2} ( z )}{2 sin ( z )}, x = \frac{y - y 1 - 4 sin ^{2} ( z )}{2 sin ( z )}; sin (z) \neq = 0

x = \frac{y + y 1 - 4 sin ^{2} ( z )}{2 sin ( z )}, x = \frac{y - y 1 - 4 sin ^{2} ( z )}{2 sin ( z )}

作图

流行的例子

(tan(x)+cos(x))/((1+sin(x)))= 1/(cos(x))

\frac{tan ( x ) + cos ( x )}{( 1 + sin ( x ) )} = \frac{1}{cos ( x )}

sin(θ)= 4/5 ,0<= θ<= pi/2

sin (θ) = \frac{4}{5}, 0 \leq θ \leq \frac{π}{2}

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\frac{( tan ( x ) + 2 sin ( x ) )}{( tan ( x ) - 2 sin ( x ) )} = 3

1 - sin (2 x) = 0

2cos(3x)+cos(2x)+1=0

2 cos (3 x) + cos (2 x) + 1 = 0