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受欢迎的三角函数 >

证明 cos^2(θ)(sec^2(θ)-1)=sin^2(θ)

解答

证明

cos^{2} (θ) (sec^{2} (θ) - 1) = sin^{2} (θ)

解答

真

求解步骤

cos^{2} (θ) (sec^{2} (θ) - 1) = sin^{2} (θ)

调整左侧

cos^{2} (θ) (sec^{2} (θ) - 1)

用 sin, cos 表示

(- 1 + sec^{2} (θ)) cos^{2} (θ)

使用基本三角恒等式:

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

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

化简

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

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

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

(\frac{1}{cos ( θ )})^{2}

使用指数法则:

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

= \frac{1 ^{2}}{cos ^{2} ( θ )}

使用法则

1^{a} = 1

1^{2} = 1

= \frac{1}{cos ^{2} ( θ )}

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

化简

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

- 1 + \frac{1}{cos ^{2} ( θ )}

将项转换为分式:

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

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

乘以:

1 \cdot cos^{2} (θ) = cos^{2} (θ)

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

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

分式相乘:

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

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

约分:

cos^{2} (θ)

= - - cos^{2} (θ) + 1

= - cos^{2} (θ) + 1

= 1 - cos^{2} (θ)

使用三角恒等式改写

1 - cos^{2} (θ)

使用毕达哥拉斯恒等式:

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

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

= sin^{2} (θ)

= sin^{2} (θ)

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 cot^2(y)sec^2(y)=1+cot^2(y)

p ro v e cot^{2} (y) sec^{2} (y) = 1 + cot^{2} (y)

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p ro v e csc (θ) - cot (θ) \cdot cos (θ) = sin (θ)

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

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

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p ro v e sin (a + b) - sin (a - b) = 2 cos (a) sin (b)