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受欢迎的三角函数 >

证明 3-4cos^2(x)=(2sin(x)+1)(2sin(x)-1)

解答

证明

3 - 4 cos^{2} (x) = (2 sin (x) + 1) (2 sin (x) - 1)

解答

真

求解步骤

3 - 4 cos^{2} (x) = (2 sin (x) + 1) (2 sin (x) - 1)

调整左侧

3 - 4 cos^{2} (x)

使用三角恒等式改写

3 - 4 cos^{2} (x)

使用毕达哥拉斯恒等式:

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

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

= 3 - 4 (1 - sin^{2} (x))

化简

3 - 4 (1 - sin^{2} (x)) : 4 sin^{2} (x) - 1

3 - 4 (1 - sin^{2} (x))

乘开

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

- 4 (1 - sin^{2} (x))

使用分配律:

a (b - c) = ab - a c

a = - 4, b = 1, c = sin^{2} (x)

= - 4 \cdot 1 - (- 4) sin^{2} (x)

使用加减运算法则

- (- a) = a

= - 4 \cdot 1 + 4 sin^{2} (x)

数字相乘:

4 \cdot 1 = 4

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

= 3 - 4 + 4 sin^{2} (x)

数字相减:

3 - 4 = - 1

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

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

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

调整右侧

(2 sin (x) + 1) (2 sin (x) - 1)

乘开

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

(- 1 + 2 sin (x)) (1 + 2 sin (x))

使用平方差公式:

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

a = 2 sin (x), b = 1

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

化简

(2 sin (x))^{2} - 1^{2} : 4 sin^{2} (x) - 1

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

使用法则

1^{a} = 1

1^{2} = 1

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

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

(2 sin (x))^{2}

使用指数法则:

(a \cdot b)^{n} = a^{n} b^{n}

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

2^{2} = 4

= 4 sin^{2} (x)

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

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

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

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 csc^{2} (θ) = (\frac{1}{sin ( θ )})^{2}

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p ro v e \frac{1 + tan ( x )}{1 + \frac{1}{t a n ( x )}} = tan (x)

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p ro v e cos (2 θ) = \frac{1}{sec ( 2 θ )}

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