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受欢迎的三角函数 >

证明 cos(3A)=4cos^3(A)-3cos(A)

解答

证明

cos (3 A) = 4 cos^{3} (A) - 3 cos (A)

解答

真

求解步骤

cos (3 A) = 4 cos^{3} (A) - 3 cos (A)

调整左侧

cos (3 A)

使用三角恒等式改写

cos (3 A)

改写为

= cos (2 A + A)

使用角和恒等式:

cos (s + t) = cos (s) cos (t) - sin (s) sin (t)

= cos (2 A) cos (A) - sin (2 A) sin (A)

使用倍角公式:

sin (2 A) = 2 sin (A) cos (A)

= cos (2 A) cos (A) - 2 sin (A) cos (A) sin (A)

化简

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

cos (2 A) cos (A) - 2 sin (A) cos (A) sin (A)

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

2 sin (A) cos (A) sin (A)

使用指数法则:

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

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

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

数字相加:

1 + 1 = 2

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

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

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

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

使用倍角公式:

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

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

使用毕达哥拉斯恒等式:

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

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

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

乘开

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

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

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

乘开

cos (A) (2 cos^{2} (A) - 1) : 2 cos^{3} (A) - cos (A)

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

使用分配律:

a (b - c) = ab - a c

a = cos (A), b = 2 cos^{2} (A), c = 1

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

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

化简

2 cos^{2} (A) cos (A) - 1 \cdot cos (A) : 2 cos^{3} (A) - cos (A)

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

2 cos^{2} (A) cos (A) = 2 cos^{3} (A)

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

使用指数法则:

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

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

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

数字相加:

2 + 1 = 3

= 2 cos^{3} (A)

1 \cdot cos (A) = cos (A)

1 cos (A)

乘以:

1 \cdot cos (A) = cos (A)

= cos (A)

= 2 cos^{3} (A) - cos (A)

= 2 cos^{3} (A) - cos (A)

= 2 cos^{3} (A) - cos (A) - 2 (1 - cos^{2} (A)) cos (A)

乘开

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

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

使用分配律:

a (b - c) = ab - a c

a = - 2 cos (A), b = 1, c = cos^{2} (A)

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

使用加减运算法则

- (- a) = a

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

化简

- 2 \cdot 1 \cdot cos (A) + 2 cos^{2} (A) cos (A) : - 2 cos (A) + 2 cos^{3} (A)

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

2 \cdot 1 \cdot cos (A) = 2 cos (A)

2 \cdot 1 cos (A)

数字相乘:

2 \cdot 1 = 2

= 2 cos (A)

2 cos^{2} (A) cos (A) = 2 cos^{3} (A)

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

使用指数法则:

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

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

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

数字相加:

2 + 1 = 3

= 2 cos^{3} (A)

= - 2 cos (A) + 2 cos^{3} (A)

= - 2 cos (A) + 2 cos^{3} (A)

= 2 cos^{3} (A) - cos (A) - 2 cos (A) + 2 cos^{3} (A)

化简

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

2 cos^{3} (A) - cos (A) - 2 cos (A) + 2 cos^{3} (A)

对同类项分组

= 2 cos^{3} (A) + 2 cos^{3} (A) - cos (A) - 2 cos (A)

同类项相加:

2 cos^{3} (A) + 2 cos^{3} (A) = 4 cos^{3} (A)

= 4 cos^{3} (A) - cos (A) - 2 cos (A)

同类项相加:

- cos (A) - 2 cos (A) = - 3 cos (A)

= 4 cos^{3} (A) - 3 cos (A)

= 4 cos^{3} (A) - 3 cos (A)

= 4 cos^{3} (A) - 3 cos (A)

我们已展示，在两侧可以有相同的形式

\Rightarrow 真

流行的例子

证明 cos(3θ)=cos^3(θ)-3sin^2(θ)cos(θ)

p ro v e cos (3 θ) = cos^{3} (θ) - 3 sin^{2} (θ) cos (θ)

证明 cos^2(B)-sin^2(B)=2cos^2(B)-1

p ro v e cos^{2} (B) - sin^{2} (B) = 2 cos^{2} (B) - 1

证明 csc^2(x)cos^2(x)= 1/(tan^2(x))

p ro v e csc^{2} (x) cos^{2} (x) = \frac{1}{tan ^{2} ( x )}

证明 1+tan^2(t)=sec^2(t)

p ro v e 1 + tan^{2} (t) = sec^{2} (t)

证明 cos(4x)=2cos^2(2x)-1

p ro v e cos (4 x) = 2 cos^{2} (2 x) - 1