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受欢迎的三角函数 >

证明 (sin(x))/(cot(x))=sec(x)-cos(x)

解答

证明

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

解答

真

求解步骤

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

调整左侧

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

用 sin, cos 表示

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

使用基本三角恒等式:

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

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

化简

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

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

使用分式法则:

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

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

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

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

使用三角恒等式改写

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

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

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

调整右侧

sec (x) - cos (x)

用 sin, cos 表示

- cos (x) + sec (x)

使用基本三角恒等式:

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

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

化简

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

- cos (x) + \frac{1}{cos ( x )}

将项转换为分式:

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

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

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{- cos ( x ) cos ( x ) + 1}{cos ( x )}

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

- cos (x) cos (x) + 1

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)

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

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

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

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

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

证明 cos(3θ)+cos(θ)=4cos^3(θ)-2cos(θ)

p ro v e cos (3 θ) + cos (θ) = 4 cos^{3} (θ) - 2 cos (θ)

证明 cos(a+b)=cos(a)cos(b)-sin(a)sin(b)

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

证明 (1+tan(θ))^2=sec^2(θ)+2tan(θ)

p ro v e (1 + tan (θ))^{2} = sec^{2} (θ) + 2 tan (θ)

证明 (1-tan(θ))^2=sec^2(θ)-2tan(θ)

p ro v e (1 - tan (θ))^{2} = sec^{2} (θ) - 2 tan (θ)