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受欢迎的三角函数 >

证明 cos^4(A)-sin^4(A)+1=2cos^2(A)

解答

证明

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

解答

真

求解步骤

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

调整左侧

cos^{4} (A) - sin^{4} (A) + 1

分解

1 - sin^{4} (A) : - (sin^{2} (A) + 1) (sin (A) + 1) (sin (A) - 1)

1 - sin^{4} (A)

因式分解出通项

- 1

= - (sin^{4} (A) - 1)

分解

sin^{4} (A) - 1 : (sin^{2} (A) + 1) (sin (A) + 1) (sin (A) - 1)

sin^{4} (A) - 1

将

sin^{4} (A) - 1

改写为

(sin^{2} (A))^{2} - 1^{2}

sin^{4} (A) - 1

将

1

改写为

1^{2}

= sin^{4} (A) - 1^{2}

使用指数法则:

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

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

= (sin^{2} (A))^{2} - 1^{2}

= (sin^{2} (A))^{2} - 1^{2}

使用平方差公式:

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

(sin^{2} (A))^{2} - 1^{2} = (sin^{2} (A) + 1) (sin^{2} (A) - 1)

= (sin^{2} (A) + 1) (sin^{2} (A) - 1)

分解

sin^{2} (A) - 1 : (sin (A) + 1) (sin (A) - 1)

sin^{2} (A) - 1

将

1

改写为

1^{2}

= sin^{2} (A) - 1^{2}

使用平方差公式:

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

sin^{2} (A) - 1^{2} = (sin (A) + 1) (sin (A) - 1)

= (sin (A) + 1) (sin (A) - 1)

= (sin^{2} (A) + 1) (sin (A) + 1) (sin (A) - 1)

= - (sin^{2} (A) + 1) (sin (A) + 1) (sin (A) - 1)

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

使用三角恒等式改写

cos^{4} (A) - (- 1 + sin (A)) (1 + sin (A)) (1 + sin^{2} (A))

使用毕达哥拉斯恒等式:

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

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

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

化简

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

乘开

(sin (A) + 1) (sin (A) - 1) : sin^{2} (A) - 1

(sin (A) + 1) (sin (A) - 1)

使用平方差公式:

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

a = sin (A), b = 1

= sin^{2} (A) - 1^{2}

使用法则

1^{a} = 1

1^{2} = 1

= sin^{2} (A) - 1

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

使用毕达哥拉斯恒等式:

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

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

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

化简

cos^{4} (A) - (- cos^{2} (A)) (2 - cos^{2} (A)) : 2 cos^{2} (A)

cos^{4} (A) - (- cos^{2} (A)) (2 - cos^{2} (A))

使用法则

- (- a) = a

= cos^{4} (A) + cos^{2} (A) (2 - cos^{2} (A))

乘开

cos^{2} (A) (2 - cos^{2} (A)) : 2 cos^{2} (A) - cos^{4} (A)

cos^{2} (A) (2 - cos^{2} (A))

使用分配律:

a (b - c) = ab - a c

a = cos^{2} (A), b = 2, c = cos^{2} (A)

= cos^{2} (A) \cdot 2 - cos^{2} (A) cos^{2} (A)

= 2 cos^{2} (A) - cos^{2} (A) cos^{2} (A)

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

cos^{2} (A) cos^{2} (A)

使用指数法则:

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

cos^{2} (A) cos^{2} (A) = cos^{2 + 2} (A)

= cos^{2 + 2} (A)

数字相加:

2 + 2 = 4

= cos^{4} (A)

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

= cos^{4} (A) + 2 cos^{2} (A) - cos^{4} (A)

同类项相加:

cos^{4} (A) - cos^{4} (A) = 0

= 2 cos^{2} (A)

= 2 cos^{2} (A)

= 2 cos^{2} (A)

我们已展示，在两侧可以有相同的形式

\Rightarrow 真

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