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受欢迎的三角函数 >

证明 sin^4(θ)-cos^4(θ)=1-2cos^2(θ)

解答

证明

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

解答

真

求解步骤

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

调整左侧

sin^{4} (θ) - cos^{4} (θ)

分解

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

- cos^{4} (θ) + sin^{4} (θ)

将

sin^{4} (θ) - cos^{4} (θ)

改写为

(sin^{2} (θ))^{2} - (cos^{2} (θ))^{2}

sin^{4} (θ) - cos^{4} (θ)

使用指数法则:

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

sin^{4} (θ) = (sin^{2} (θ))^{2}

= (sin^{2} (θ))^{2} - cos^{4} (θ)

使用指数法则:

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

cos^{4} (θ) = (cos^{2} (θ))^{2}

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

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

使用平方差公式:

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

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

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

分解

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

sin^{2} (θ) - cos^{2} (θ)

使用平方差公式:

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

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

= (sin (θ) + cos (θ)) (sin (θ) - cos (θ))

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

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

使用三角恒等式改写

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

乘开

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

(sin (θ) + cos (θ)) (sin (θ) - cos (θ))

使用平方差公式:

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

a = sin (θ), b = cos (θ)

= sin^{2} (θ) - cos^{2} (θ)

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

使用倍角公式:

cos^{2} (θ) - sin^{2} (θ) = cos (2 θ)

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

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

化简

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

使用倍角公式:

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

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

使用毕达哥拉斯恒等式:

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

= - (2 cos^{2} (θ) - 1) \cdot 1

化简

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

- (2 cos^{2} (θ) - 1) \cdot 1

乘以:

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

= - (2 cos^{2} (θ) - 1)

打开括号

= - (2 cos^{2} (θ)) - (- 1)

使用加减运算法则

- (- a) = a, - (a) = - a

= - 2 cos^{2} (θ) + 1

= - 2 cos^{2} (θ) + 1

= - 2 cos^{2} (θ) + 1

= 1 - 2 cos^{2} (θ)

我们已展示，在两侧可以有相同的形式

\Rightarrow 真

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