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受欢迎的三角函数 >

sin^2(θ)+cos^2(θ)=2

解答

sin^{2} (θ) + cos^{2} (θ) = 2

解答

θ \in R 无解

求解步骤

sin^{2} (θ) + cos^{2} (θ) = 2

两边减去

cos^{2} (θ)

sin^{2} (θ) = 2 - cos^{2} (θ)

两边进行平方

(sin^{2} (θ))^{2} = (2 - cos^{2} (θ))^{2}

两边减去

(2 - cos^{2} (θ))^{2}

sin^{4} (θ) - 4 + 4 cos^{2} (θ) - cos^{4} (θ) = 0

使用指数法则:

a^{b} = a^{2} a^{b - 2}

- 4 - cos^{4} (θ) + 4 cos^{2} (θ) + sin^{2} (θ) sin^{2} (θ) = 0

使用三角恒等式改写

- 4 - cos^{4} (θ) + 4 cos^{2} (θ) + sin^{2} (θ) sin^{2} (θ)

使用毕达哥拉斯恒等式:

cos^{2} (x) + sin^{2} (x) = 1

sin^{2} (x) = 1 - cos^{2} (x)

= - 4 - cos^{4} (θ) + 4 cos^{2} (θ) + (1 - cos^{2} (θ)) (1 - cos^{2} (θ))

化简

- 4 - cos^{4} (θ) + 4 cos^{2} (θ) + (1 - cos^{2} (θ)) (1 - cos^{2} (θ)) : 2 cos^{2} (θ) - 3

- 4 - cos^{4} (θ) + 4 cos^{2} (θ) + (1 - cos^{2} (θ)) (1 - cos^{2} (θ))

(1 - cos^{2} (θ)) (1 - cos^{2} (θ)) = (1 - cos^{2} (θ))^{2}

(1 - cos^{2} (θ)) (1 - cos^{2} (θ))

使用指数法则:

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

(1 - cos^{2} (θ)) (1 - cos^{2} (θ)) = (1 - cos^{2} (θ))^{1 + 1}

= (1 - cos^{2} (θ))^{1 + 1}

数字相加:

1 + 1 = 2

= (1 - cos^{2} (θ))^{2}

= - 4 - cos^{4} (θ) + 4 cos^{2} (θ) + (- cos^{2} (θ) + 1)^{2}

(1 - cos^{2} (θ))^{2} : 1 - 2 cos^{2} (θ) + cos^{4} (θ)

使用完全平方公式:

(a - b)^{2} = a^{2} - 2 ab + b^{2}

a = 1, b = cos^{2} (θ)

= 1^{2} - 2 \cdot 1 \cdot cos^{2} (θ) + (cos^{2} (θ))^{2}

化简

1^{2} - 2 \cdot 1 \cdot cos^{2} (θ) + (cos^{2} (θ))^{2} : 1 - 2 cos^{2} (θ) + cos^{4} (θ)

1^{2} - 2 \cdot 1 \cdot cos^{2} (θ) + (cos^{2} (θ))^{2}

使用法则

1^{a} = 1

1^{2} = 1

= 1 - 2 \cdot 1 \cdot cos^{2} (θ) + (cos^{2} (θ))^{2}

2 \cdot 1 \cdot cos^{2} (θ) = 2 cos^{2} (θ)

2 \cdot 1 \cdot cos^{2} (θ)

数字相乘:

2 \cdot 1 = 2

= 2 cos^{2} (θ)

(cos^{2} (θ))^{2} = cos^{4} (θ)

(cos^{2} (θ))^{2}

使用指数法则:

(a^{b})^{c} = a^{b c}

= cos^{2 \cdot 2} (θ)

数字相乘:

2 \cdot 2 = 4

= cos^{4} (θ)

= 1 - 2 cos^{2} (θ) + cos^{4} (θ)

= 1 - 2 cos^{2} (θ) + cos^{4} (θ)

= - 4 - cos^{4} (θ) + 4 cos^{2} (θ) + 1 - 2 cos^{2} (θ) + cos^{4} (θ)

化简

- 4 - cos^{4} (θ) + 4 cos^{2} (θ) + 1 - 2 cos^{2} (θ) + cos^{4} (θ) : 2 cos^{2} (θ) - 3

- 4 - cos^{4} (θ) + 4 cos^{2} (θ) + 1 - 2 cos^{2} (θ) + cos^{4} (θ)

对同类项分组

= - cos^{4} (θ) + 4 cos^{2} (θ) - 2 cos^{2} (θ) + cos^{4} (θ) - 4 + 1

同类项相加:

4 cos^{2} (θ) - 2 cos^{2} (θ) = 2 cos^{2} (θ)

= - cos^{4} (θ) + 2 cos^{2} (θ) + cos^{4} (θ) - 4 + 1

同类项相加:

- cos^{4} (θ) + cos^{4} (θ) = 0

= 2 cos^{2} (θ) - 4 + 1

数字相加/相减:

- 4 + 1 = - 3

= 2 cos^{2} (θ) - 3

= 2 cos^{2} (θ) - 3

= 2 cos^{2} (θ) - 3

- 3 + 2 cos^{2} (θ) = 0

用替代法求解

- 3 + 2 cos^{2} (θ) = 0

令

: cos (θ) = u

- 3 + 2 u^{2} = 0

- 3 + 2 u^{2} = 0 : u = \frac{3}{2}, u = - \frac{3}{2}

- 3 + 2 u^{2} = 0

将

3

到右边

- 3 + 2 u^{2} = 0

两边加上

3

- 3 + 2 u^{2} + 3 = 0 + 3

化简

2 u^{2} = 3

2 u^{2} = 3

两边除以

2

2 u^{2} = 3

两边除以

2

\frac{2 u ^{2}}{2} = \frac{3}{2}

化简

u^{2} = \frac{3}{2}

u^{2} = \frac{3}{2}

对于

x^{2} = f (a)

解为

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

u = \frac{3}{2}, u = - \frac{3}{2}

u = cos (θ)

代回

cos (θ) = \frac{3}{2}, cos (θ) = - \frac{3}{2}

cos (θ) = \frac{3}{2}, cos (θ) = - \frac{3}{2}

cos (θ) = \frac{3}{2} :

无解

cos (θ) = \frac{3}{2}

- 1 \leq cos (x) \leq 1

无解

cos (θ) = - \frac{3}{2} :

无解

cos (θ) = - \frac{3}{2}

- 1 \leq cos (x) \leq 1

无解

合并所有解

无解

将解代入原方程进行验证

将它们代入

sin^{2} (θ) + cos^{2} (θ) = 2

检验解是否符合

去除与方程不符的解。

θ \in R 无解

作图

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