zs

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

受欢迎的三角函数 >

证明 (tan^2(θ))/(sin^2(θ))-1=tan^2(θ)

解答

证明

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

解答

真

求解步骤

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

调整左侧

\frac{tan ^{2} ( θ )}{sin ^{2} ( θ )} - 1

用 sin, cos 表示

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

使用基本三角恒等式:

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

= - 1 + \frac{( \frac{s i n ( θ )}{c o s ( θ )} ) ^{2}}{sin ^{2} ( θ )}

化简

- 1 + \frac{( \frac{s i n ( θ )}{c o s ( θ )} ) ^{2}}{sin ^{2} ( θ )} : \frac{- cos ^{2} ( θ ) + 1}{cos ^{2} ( θ )}

- 1 + \frac{( \frac{s i n ( θ )}{c o s ( θ )} ) ^{2}}{sin ^{2} ( θ )}

\frac{( \frac{s i n ( θ )}{c o s ( θ )} ) ^{2}}{sin ^{2} ( θ )} = \frac{1}{cos ^{2} ( θ )}

\frac{( \frac{s i n ( θ )}{c o s ( θ )} ) ^{2}}{sin ^{2} ( θ )}

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

(\frac{sin ( θ )}{cos ( θ )})^{2}

使用指数法则:

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

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

= \frac{\frac{s i n ^{2} ( θ )}{c o s ^{2} ( θ )}}{sin ^{2} ( θ )}

使用分式法则:

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

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

约分:

sin^{2} (θ)

= \frac{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 )}

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

使用三角恒等式改写

\frac{1 - cos ^{2} ( θ )}{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)

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

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

使用三角恒等式改写

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

使用基本三角恒等式:

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

\frac{sin ( θ ) tan ( θ )}{cos ( θ )}

= tan (θ) \frac{sin ( θ )}{cos ( θ )}

使用基本三角恒等式:

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

tan (θ) tan (θ)

化简

tan (θ) tan (θ) : tan^{2} (θ)

tan (θ) tan (θ)

使用指数法则:

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

tan (θ) tan (θ) = tan^{1 + 1} (θ)

= tan^{1 + 1} (θ)

数字相加:

1 + 1 = 2

= tan^{2} (θ)

tan^{2} (θ)

tan^{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 csc(θ)cos^2(θ)+sin(θ)=csc(θ)

p ro v e csc (θ) cos^{2} (θ) + sin (θ) = csc (θ)

证明 sec(θ+pi/2)=csc(θ)

p ro v e sec (θ + \frac{π}{2}) = csc (θ)

证明 (1+csc(θ))/(sec(θ))-cot(θ)=cos(θ)

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

证明 sin(2θ)=(2cot(θ))/(1+cot^2(θ))

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

证明 tan(θ)=(1-cos(2θ))/(sin(2θ))

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