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受欢迎的三角函数 >

证明 1-cos^2(x)=(tan^2(x))/(sec^2(x))

解答

证明

1 - cos^{2} (x) = \frac{tan ^{2} ( x )}{sec ^{2} ( x )}

解答

真

求解步骤

1 - cos^{2} (x) = \frac{tan ^{2} ( x )}{sec ^{2} ( x )}

调整右侧

\frac{tan ^{2} ( x )}{sec ^{2} ( x )}

用 sin, cos 表示

\frac{tan ^{2} ( x )}{sec ^{2} ( x )}

使用基本三角恒等式:

tan (x) = \frac{sin ( x )}{cos ( x )}

= \frac{( \frac{s i n ( x )}{c o s ( x )} ) ^{2}}{sec ^{2} ( x )}

使用基本三角恒等式:

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

= \frac{( \frac{s i n ( x )}{c o s ( x )} ) ^{2}}{( \frac{1}{c o s ( x )} ) ^{2}}

化简

\frac{( \frac{s i n ( x )}{c o s ( x )} ) ^{2}}{( \frac{1}{c o s ( x )} ) ^{2}} : sin^{2} (x)

\frac{( \frac{s i n ( x )}{c o s ( x )} ) ^{2}}{( \frac{1}{c o s ( x )} ) ^{2}}

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

(\frac{1}{cos ( x )})^{2}

使用指数法则:

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

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

使用法则

1^{a} = 1

1^{2} = 1

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

= \frac{( \frac{s i n ( x )}{c o s ( x )} ) ^{2}}{\frac{1}{c o s ^{2} ( x )}}

(\frac{sin ( x )}{cos ( x )})^{2} = \frac{sin ^{2} ( x )}{cos ^{2} ( x )}

(\frac{sin ( x )}{cos ( x )})^{2}

使用指数法则:

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

= \frac{sin ^{2} ( x )}{cos ^{2} ( x )}

= \frac{\frac{s i n ^{2} ( x )}{c o s ^{2} ( x )}}{\frac{1}{c o s ^{2} ( x )}}

分式相除:

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

= \frac{sin ^{2} ( x ) cos ^{2} ( x )}{cos ^{2} ( x ) \cdot 1}

整理后得

= \frac{sin ^{2} ( x ) cos ^{2} ( x )}{cos ^{2} ( x )}

约分:

cos^{2} (x)

= sin^{2} (x)

= sin^{2} (x)

= sin^{2} (x)

使用三角恒等式改写

sin^{2} (x)

使用毕达哥拉斯恒等式:

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

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

= 1 - cos^{2} (x)

= 1 - cos^{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

证明 ((cos^2(x)))/((1-sin(x)))=1+sin(x)

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

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p ro v e 8 csc^{2} (x) - 3 cot^{2} (x) = 3 + 5 csc^{2} (x)

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