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受欢迎的三角函数 >

证明 (1+sec(θ))(1-cos(θ))=sec(θ)-cos(θ)

解答

证明

(1 + sec (θ)) (1 - cos (θ)) = sec (θ) - cos (θ)

解答

真

求解步骤

(1 + sec (θ)) (1 - cos (θ)) = sec (θ) - cos (θ)

调整左侧

(1 + sec (θ)) (1 - cos (θ))

用 sin, cos 表示

(1 + sec (θ)) (1 - cos (θ))

使用基本三角恒等式:

sec (x) = \frac{1}{cos ( x )}

= (1 + \frac{1}{cos ( θ )}) (1 - cos (θ))

化简

(1 + \frac{1}{cos ( θ )}) (1 - cos (θ)) : \frac{( cos ( θ ) + 1 ) ( 1 - cos ( θ ) )}{cos ( θ )}

(1 + \frac{1}{cos ( θ )}) (1 - cos (θ))

化简

1 + \frac{1}{cos ( θ )} : \frac{cos ( θ ) + 1}{cos ( θ )}

1 + \frac{1}{cos ( θ )}

将项转换为分式:

1 = \frac{1 cos ( θ )}{cos ( θ )}

= \frac{1 \cdot cos ( θ )}{cos ( θ )} + \frac{1}{cos ( θ )}

因为分母相等，所以合并分式:

\frac{a}{c} \pm \frac{b}{c} = \frac{a \pm b}{c}

= \frac{1 \cdot cos ( θ ) + 1}{cos ( θ )}

乘以:

1 \cdot cos (θ) = cos (θ)

= \frac{cos ( θ ) + 1}{cos ( θ )}

= \frac{cos ( θ ) + 1}{cos ( θ )} (- cos (θ) + 1)

分式相乘:

a \cdot \frac{b}{c} = \frac{a \cdot b}{c}

= \frac{( cos ( θ ) + 1 ) ( 1 - cos ( θ ) )}{cos ( θ )}

= \frac{( cos ( θ ) + 1 ) ( 1 - cos ( θ ) )}{cos ( θ )}

= \frac{( 1 + cos ( θ ) ) ( 1 - cos ( θ ) )}{cos ( θ )}

乘开

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

(1 + cos (θ)) (1 - cos (θ))

使用平方差公式:

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

a = 1, b = cos (θ)

= 1^{2} - cos^{2} (θ)

使用法则

1^{a} = 1

1^{2} = 1

= 1 - cos^{2} (θ)

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

调整右侧

sec (θ) - cos (θ)

用 sin, cos 表示

- cos (θ) + sec (θ)

使用基本三角恒等式:

sec (x) = \frac{1}{cos ( x )}

= - cos (θ) + \frac{1}{cos ( θ )}

化简

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

- cos (θ) + \frac{1}{cos ( θ )}

将项转换为分式:

cos (θ) = \frac{cos ( θ ) cos ( θ )}{cos ( θ )}

= - \frac{cos ( θ ) cos ( θ )}{cos ( θ )} + \frac{1}{cos ( θ )}

因为分母相等，所以合并分式:

\frac{a}{c} \pm \frac{b}{c} = \frac{a \pm b}{c}

= \frac{- cos ( θ ) cos ( θ ) + 1}{cos ( θ )}

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

- cos (θ) cos (θ) + 1

cos (θ) cos (θ) = cos^{2} (θ)

cos (θ) cos (θ)

使用指数法则:

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

cos (θ) cos (θ) = cos^{1 + 1} (θ)

= cos^{1 + 1} (θ)

数字相加:

1 + 1 = 2

= cos^{2} (θ)

= - cos^{2} (θ) + 1

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

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

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

我们已展示，在两侧可以有相同的形式

\Rightarrow 真

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