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受欢迎的三角函数 >

证明 (csc(A))/(csc(A)-sin(A))=sec^2(A)

解答

证明

\frac{csc ( A )}{csc ( A ) - sin ( A )} = sec^{2} (A)

解答

真

求解步骤

\frac{csc ( A )}{csc ( A ) - sin ( A )} = sec^{2} (A)

调整左侧

\frac{csc ( A )}{csc ( A ) - sin ( A )}

用 sin, cos 表示

\frac{csc ( A )}{csc ( A ) - sin ( A )}

使用基本三角恒等式:

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

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

化简

\frac{\frac{1}{s i n ( A )}}{\frac{1}{s i n ( A )} - sin ( A )} : \frac{1}{1 - sin ^{2} ( A )}

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

使用分式法则:

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

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

化简

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

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

将项转换为分式:

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

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

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 ( A ) sin ( A )}{sin ( A )}

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

1 - sin (A) sin (A)

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

sin (A) sin (A)

使用指数法则:

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

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

= sin^{1 + 1} (A)

数字相加:

1 + 1 = 2

= sin^{2} (A)

= 1 - sin^{2} (A)

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

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

乘

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

sin (A) \frac{1 - sin ^{2} ( A )}{sin ( A )}

分式相乘:

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

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

约分:

sin (A)

= 1 - sin^{2} (A)

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

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

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

调整右侧

sec^{2} (A)

用 sin, cos 表示

sec^{2} (A)

使用基本三角恒等式:

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

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

化简

(\frac{1}{cos ( A )})^{2} : \frac{1}{cos ^{2} ( A )}

(\frac{1}{cos ( A )})^{2}

使用指数法则:

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

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

使用法则

1^{a} = 1

1^{2} = 1

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

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

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

使用三角恒等式改写

\frac{1}{cos ^{2} ( A )}

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}{1 - sin ^{2} ( A )}

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

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^4(x)(cot^2(x))=csc^4(x)

p ro v e csc^{4} (x) (cot^{2} (x)) = csc^{4} (x)

证明 (sin^2(x))/(cos^2(x))+1=sec^2(x)

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

证明 cos^2(x)tan^2(2x)=sin^2(x)

p ro v e cos^{2} (x) tan^{2} (2 x) = sin^{2} (x)

证明 (sin(2x))/2 =(tan(x))/(1+tan^2(x))

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

证明 csc(pi/(12))=sec((5pi)/(12))

p ro v e csc (\frac{π}{12}) = sec (\frac{5 π}{12})