zs

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

受欢迎的三角函数 >

证明 4cos^3(a)-3cos(a)=cos(3a)

解答

证明

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

解答

真

求解步骤

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

调整右侧

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 sin^2(θ)+(1/2)^2=1

p ro v e sin^{2} (θ) + (\frac{1}{2})^{2} = 1

证明 sin(θ)(cot(θ)+tan(θ))=tan(θ)

p ro v e sin (θ) (cot (θ) + tan (θ)) = tan (θ)

证明-csc^2(x)=(sec(x))/(2sin(x))

p ro v e - csc^{2} (x) = \frac{sec ( x )}{2 sin ( x )}

证明 (sec(x)cos(x))/(cot(x))=tan(x)

p ro v e \frac{sec ( x ) cos ( x )}{cot ( x )} = tan (x)

证明 2sin^2((9θ)/2)=1-cos(9θ)

p ro v e 2 sin^{2} (\frac{9 θ}{2}) = 1 - cos (9 θ)