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受欢迎的三角函数 >

证明 (csc(x))/(sin(x))-(cot(x))/(tan(x))=1

解答

证明

\frac{csc ( x )}{sin ( x )} - \frac{cot ( x )}{tan ( x )} = 1

解答

真

求解步骤

\frac{csc ( x )}{sin ( x )} - \frac{cot ( x )}{tan ( x )} = 1

调整左侧

\frac{csc ( x )}{sin ( x )} - \frac{cot ( x )}{tan ( x )}

用 sin, cos 表示

- \frac{cot ( x )}{tan ( x )} + \frac{csc ( x )}{sin ( x )}

使用基本三角恒等式:

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

= - \frac{\frac{c o s ( x )}{s i n ( x )}}{tan ( x )} + \frac{csc ( x )}{sin ( x )}

使用基本三角恒等式:

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

= - \frac{\frac{c o s ( x )}{s i n ( x )}}{\frac{s i n ( x )}{c o s ( x )}} + \frac{csc ( x )}{sin ( x )}

使用基本三角恒等式:

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

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

化简

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

- \frac{\frac{c o s ( x )}{s i n ( x )}}{\frac{s i n ( x )}{c o s ( x )}} + \frac{\frac{1}{s i n ( x )}}{sin ( x )}

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

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

分式相除:

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

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

cos (x) cos (x) = cos^{2} (x)

cos (x) cos (x)

使用指数法则:

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

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

= cos^{1 + 1} (x)

数字相加:

1 + 1 = 2

= cos^{2} (x)

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

sin (x) sin (x) = sin^{2} (x)

sin (x) sin (x)

使用指数法则:

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

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

= sin^{1 + 1} (x)

数字相加:

1 + 1 = 2

= sin^{2} (x)

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

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

\frac{\frac{1}{s i n ( x )}}{sin ( x )}

使用分式法则:

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

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

sin (x) sin (x) = sin^{2} (x)

sin (x) sin (x)

使用指数法则:

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

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

= sin^{1 + 1} (x)

数字相加:

1 + 1 = 2

= sin^{2} (x)

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

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

使用法则

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

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

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

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

使用三角恒等式改写

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

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} ( x )}{sin ^{2} ( x )}

使用法则

\frac{a}{a} = 1

= 1

= 1

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(x)-cot(x))/(sec(x)-1)=cot(x)

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

证明 (1+cos(x))/(1-cos(x))-(1-cos(x))/(1+cos(x))=4cot(x)csc(x)

p ro v e \frac{1 + cos ( x )}{1 - cos ( x )} - \frac{1 - cos ( x )}{1 + cos ( x )} = 4 cot (x) csc (x)

证明 cos(z+w)=cos(z)cos(w)-sin(z)sin(w)

p ro v e cos (z + w) = cos (z) cos (w) - sin (z) sin (w)

证明 tan(pi-θ)=-tan(θ)

p ro v e tan (π - θ) = - tan (θ)

证明 (sin(θ))/(1-cos(θ))=csc(θ)+cot(θ)

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