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受欢迎的三角函数 >

证明 csc(x)(csc(x)+sin(-x))=cot^2(x)

解答

证明

csc (x) (csc (x) + sin (- x)) = cot^{2} (x)

解答

真

求解步骤

csc (x) (csc (x) + sin (- x)) = cot^{2} (x)

调整左侧

csc (x) (csc (x) + sin (- x))

使用负角恒等式:

sin (- x) = - sin (x)

= csc (x) (csc (x) - sin (x))

用 sin, cos 表示

(csc (x) - sin (x)) csc (x)

使用基本三角恒等式:

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

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

化简

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

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

分式相乘:

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

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

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

1 \cdot (\frac{1}{sin ( x )} - sin (x))

乘以:

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

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

去除括号:

(a) = a

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

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

化简

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

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

将项转换为分式:

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

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

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

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)

= 1 - sin^{2} (x)

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

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

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

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

调整右侧

cot^{2} (x)

用 sin, cos 表示

cot^{2} (x)

使用基本三角恒等式:

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

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

化简

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

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

使用指数法则:

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

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

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

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

使用三角恒等式改写

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

使用毕达哥拉斯恒等式:

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

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

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

= \frac{1 - sin ^{2} ( x )}{sin ^{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 tan(x)+cot(x)=(sec(x))*(csc(x))

p ro v e tan (x) + cot (x) = (sec (x)) \cdot (csc (x))

证明 (csc(θ)sin(θ))/(cos(θ))=sec(θ)

p ro v e \frac{csc ( θ ) sin ( θ )}{cos ( θ )} = sec (θ)

证明 cos(-pi/4)= 1/(sqrt(2))

p ro v e cos (- \frac{π}{4}) = \frac{1}{2}

证明 (tan(x)+cot(x))/(sec(x)csc(x))=1

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

证明 tan^2(a)+sin^2(a)=tan^2(a)sin^2(a)

p ro v e tan^{2} (a) + sin^{2} (a) = tan^{2} (a) sin^{2} (a)