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受欢迎的三角函数 >

证明 csc^2(t)tan^2(t)-1=tan^2(t)

解答

证明

csc^{2} (t) tan^{2} (t) - 1 = tan^{2} (t)

解答

真

求解步骤

csc^{2} (t) tan^{2} (t) - 1 = tan^{2} (t)

调整左侧

csc^{2} (t) tan^{2} (t) - 1

用 sin, cos 表示

- 1 + csc^{2} (t) tan^{2} (t)

使用基本三角恒等式:

csc (x) = \frac{1}{sin ( x )}

= - 1 + (\frac{1}{sin ( t )})^{2} tan^{2} (t)

使用基本三角恒等式:

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

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

化简

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

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

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

(\frac{1}{sin ( t )})^{2} (\frac{sin ( t )}{cos ( t )})^{2}

(\frac{1}{sin ( t )})^{2} = \frac{1}{sin ^{2} ( t )}

(\frac{1}{sin ( t )})^{2}

使用指数法则:

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

= \frac{1 ^{2}}{sin ^{2} ( t )}

使用法则

1^{a} = 1

1^{2} = 1

= \frac{1}{sin ^{2} ( t )}

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

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

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

使用指数法则:

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

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

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

分式相乘:

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

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

约分:

sin^{2} (t)

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

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

将项转换为分式:

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

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

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} ( t ) + 1}{cos ^{2} ( t )}

乘以:

1 \cdot cos^{2} (t) = cos^{2} (t)

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

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

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

使用三角恒等式改写

\frac{1 - cos ^{2} ( t )}{cos ^{2} ( t )}

使用毕达哥拉斯恒等式:

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

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

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

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

使用三角恒等式改写

= \frac{sin ( t )}{cos ( t )} \cdot \frac{sin ( t )}{cos ( t )}

使用基本三角恒等式:

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

\frac{sin ( t ) tan ( t )}{cos ( t )}

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

使用基本三角恒等式:

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

tan (t) tan (t)

化简

tan (t) tan (t) : tan^{2} (t)

tan (t) tan (t)

使用指数法则:

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

tan (t) tan (t) = tan^{1 + 1} (t)

= tan^{1 + 1} (t)

数字相加:

1 + 1 = 2

= tan^{2} (t)

tan^{2} (t)

tan^{2} (t)

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 1/2 cos(x)=(cos(x))/2

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

证明 (sec^2(θ)-1)/(sin^2(θ))=sec^2(θ)

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

证明 sin(cot(x)+tan(x))=sec(x)

p ro v e sin (cot (x) + tan (x)) = sec (x)

证明 sin^2(2x)=4sin^2(x)*cos^2(x)

p ro v e sin^{2} (2 x) = 4 sin^{2} (x) \cdot cos^{2} (x)

证明 (sec(x)csc(x))/2 =csc(2x)

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