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受欢迎的三角函数 >

441/841+sin^2(t)=1

解答

\frac{441}{841} + sin^{2} (t) = 1

解答

t = 0.76101 \dots + 2 πn, t = π - 0.76101 \dots + 2 πn, t = - 0.76101 \dots + 2 πn, t = π + 0.76101 \dots + 2 πn

+1

度数

t = 43.60281 \dots^{\circ} + 36 0^{\circ} n, t = 136.39718 \dots^{\circ} + 36 0^{\circ} n, t = - 43.60281 \dots^{\circ} + 36 0^{\circ} n, t = 223.60281 \dots^{\circ} + 36 0^{\circ} n

求解步骤

\frac{441}{841} + sin^{2} (t) = 1

用替代法求解

\frac{441}{841} + sin^{2} (t) = 1

令

: sin (t) = u

\frac{441}{841} + u^{2} = 1

\frac{441}{841} + u^{2} = 1 : u = \frac{20}{29}, u = - \frac{20}{29}

\frac{441}{841} + u^{2} = 1

将

\frac{441}{841}

到右边

\frac{441}{841} + u^{2} = 1

两边减去

\frac{441}{841}

\frac{441}{841} + u^{2} - \frac{441}{841} = 1 - \frac{441}{841}

化简

u^{2} = 1 - \frac{441}{841}

u^{2} = 1 - \frac{441}{841}

化简

1 - \frac{441}{841} : \frac{400}{841}

1 - \frac{441}{841}

将项转换为分式:

1 = \frac{1 \cdot 841}{841}

= \frac{1 \cdot 841}{841} - \frac{441}{841}

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{1 \cdot 841 - 441}{841}

1 \cdot 841 - 441 = 400

1 \cdot 841 - 441

数字相乘:

1 \cdot 841 = 841

= 841 - 441

数字相减:

841 - 441 = 400

= 400

= \frac{400}{841}

对于

x^{2} = f (a)

解为

x = f (a), - f (a)

u = \frac{400}{841}, u = - \frac{400}{841}

\frac{400}{841} = \frac{20}{29}

\frac{400}{841}

使用根式运算法则:

n \frac{a}{b} = \frac{n a}{n b},

假定

a \geq 0, b \geq 0

= \frac{400}{841}

841 = 29

841

因式分解数字:

841 = 2 9^{2}

= 2 9^{2}

使用根式运算法则:

n a^{n} = a

2 9^{2} = 29

= 29

= \frac{400}{29}

400 = 20

400

因式分解数字:

400 = 2 0^{2}

= 2 0^{2}

使用根式运算法则:

n a^{n} = a

2 0^{2} = 20

= 20

= \frac{20}{29}

- \frac{400}{841} = - \frac{20}{29}

- \frac{400}{841}

化简

\frac{400}{841} : \frac{20}{29}

\frac{400}{841}

使用根式运算法则:

n \frac{a}{b} = \frac{n a}{n b},

假定

a \geq 0, b \geq 0

= \frac{400}{841}

841 = 29

841

因式分解数字:

841 = 2 9^{2}

= 2 9^{2}

使用根式运算法则:

n a^{n} = a

2 9^{2} = 29

= 29

= \frac{400}{29}

400 = 20

400

因式分解数字:

400 = 2 0^{2}

= 2 0^{2}

使用根式运算法则:

n a^{n} = a

2 0^{2} = 20

= 20

= \frac{20}{29}

= - \frac{20}{29}

u = \frac{20}{29}, u = - \frac{20}{29}

u = sin (t)

代回

sin (t) = \frac{20}{29}, sin (t) = - \frac{20}{29}

sin (t) = \frac{20}{29}, sin (t) = - \frac{20}{29}

sin (t) = \frac{20}{29} : t = arcsin (\frac{20}{29}) + 2 πn, t = π - arcsin (\frac{20}{29}) + 2 πn

sin (t) = \frac{20}{29}

使用反三角函数性质

sin (t) = \frac{20}{29}

sin (t) = \frac{20}{29}

的通解

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

t = arcsin (\frac{20}{29}) + 2 πn, t = π - arcsin (\frac{20}{29}) + 2 πn

t = arcsin (\frac{20}{29}) + 2 πn, t = π - arcsin (\frac{20}{29}) + 2 πn

sin (t) = - \frac{20}{29} : t = arcsin (- \frac{20}{29}) + 2 πn, t = π + arcsin (\frac{20}{29}) + 2 πn

sin (t) = - \frac{20}{29}

使用反三角函数性质

sin (t) = - \frac{20}{29}

sin (t) = - \frac{20}{29}

的通解

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

t = arcsin (- \frac{20}{29}) + 2 πn, t = π + arcsin (\frac{20}{29}) + 2 πn

t = arcsin (- \frac{20}{29}) + 2 πn, t = π + arcsin (\frac{20}{29}) + 2 πn

合并所有解

t = arcsin (\frac{20}{29}) + 2 πn, t = π - arcsin (\frac{20}{29}) + 2 πn, t = arcsin (- \frac{20}{29}) + 2 πn, t = π + arcsin (\frac{20}{29}) + 2 πn

以小数形式表示解

t = 0.76101 \dots + 2 πn, t = π - 0.76101 \dots + 2 πn, t = - 0.76101 \dots + 2 πn, t = π + 0.76101 \dots + 2 πn

作图

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