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受欢迎的三角函数 >

证明 cos^2(2x)+4sin^2(x)cos^2(x)=1

解答

证明

cos^{2} (2 x) + 4 sin^{2} (x) cos^{2} (x) = 1

解答

真

求解步骤

cos^{2} (2 x) + 4 sin^{2} (x) cos^{2} (x) = 1

调整左侧

cos^{2} (2 x) + 4 sin^{2} (x) cos^{2} (x)

使用三角恒等式改写

cos^{2} (2 x) + 4 sin^{2} (x) cos^{2} (x)

使用倍角公式:

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

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

化简

(cos^{2} (x) - sin^{2} (x))^{2} + 4 sin^{2} (x) cos^{2} (x) : cos^{4} (x) + 2 sin^{2} (x) cos^{2} (x) + sin^{4} (x)

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

(cos^{2} (x) - sin^{2} (x))^{2} : cos^{4} (x) - 2 cos^{2} (x) sin^{2} (x) + sin^{4} (x)

使用完全平方公式:

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

a = cos^{2} (x), b = sin^{2} (x)

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

化简

(cos^{2} (x))^{2} - 2 cos^{2} (x) sin^{2} (x) + (sin^{2} (x))^{2} : cos^{4} (x) - 2 cos^{2} (x) sin^{2} (x) + sin^{4} (x)

(cos^{2} (x))^{2} - 2 cos^{2} (x) sin^{2} (x) + (sin^{2} (x))^{2}

(cos^{2} (x))^{2} = cos^{4} (x)

(cos^{2} (x))^{2}

使用指数法则:

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

= cos^{2 \cdot 2} (x)

数字相乘:

2 \cdot 2 = 4

= cos^{4} (x)

(sin^{2} (x))^{2} = sin^{4} (x)

(sin^{2} (x))^{2}

使用指数法则:

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

= sin^{2 \cdot 2} (x)

数字相乘:

2 \cdot 2 = 4

= sin^{4} (x)

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

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

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

同类项相加:

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

= cos^{4} (x) + 2 sin^{2} (x) cos^{2} (x) + sin^{4} (x)

= cos^{4} (x) + 2 sin^{2} (x) cos^{2} (x) + sin^{4} (x)

= cos^{4} (x) + 2 sin^{2} (x) cos^{2} (x) + sin^{4} (x)

分解

cos^{4} (x) + sin^{4} (x) + 2 cos^{2} (x) sin^{2} (x) : (sin^{2} (x) + cos^{2} (x))^{2}

cos^{4} (x) + sin^{4} (x) + 2 cos^{2} (x) sin^{2} (x)

将表达式拆分成组

sin^{4} (x) + 2 sin^{2} (x) cos^{2} (x) + cos^{4} (x)

改写为形式

a x^{2} + b x y + c y^{2}

Let

sin^{2} (x) = sin (x), cos^{2} (x) = cos (x)

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

定义

1

的因数:

1

1

约数 (因数)

找到

1

的质因数:

加 1

1

1

的因数

1

对于每两个因数

u * v = 1

，检验是否

u + v = 2

检验

u = 1, v = 1 : u * v = 1, u + v = 2 \Rightarrow

真

u = 1, v = 1

分组为

(a x^{2} + ux y^{2}) + (v x^{2} y + c y^{2})

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

重设

sin (x) = sin^{2} (x), cos (x) = cos^{2} (x)

(sin^{4} (x) + sin^{2} (x) cos^{2} (x)) + (sin^{2} (x) cos^{2} (x) + cos^{4} (x))

= (sin^{4} (x) + sin^{2} (x) cos^{2} (x)) + (sin^{2} (x) cos^{2} (x) + cos^{4} (x))

从

sin^{4} (x) + sin^{2} (x) cos^{2} (x)

分解出因式

sin^{2} (x)

:

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

sin^{4} (x) + sin^{2} (x) cos^{2} (x)

使用指数法则:

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

sin^{4} (x) = sin^{2} (x) sin^{2} (x)

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

因式分解出通项

sin^{2} (x)

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

从

sin^{2} (x) cos^{2} (x) + cos^{4} (x)

分解出因式

cos^{2} (x)

:

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

sin^{2} (x) cos^{2} (x) + cos^{4} (x)

使用指数法则:

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

cos^{4} (x) = cos^{2} (x) cos^{2} (x)

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

因式分解出通项

cos^{2} (x)

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

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

因式分解出通项

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

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

整理后得

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

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

使用三角恒等式改写

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

使用毕达哥拉斯恒等式:

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

= 1^{2}

使用法则

1^{a} = 1

= 1

= 1

我们已展示，在两侧可以有相同的形式

\Rightarrow 真

流行的例子

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p ro v e cos (x) (sec (x) - 1) = 1 - cos (x)

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p ro v e csc^{2} (θ) (1 - cos^{2} (θ)) = - cos (\frac{π}{2})

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