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受欢迎的三角函数 >

证明 (csc(θ))/(sin(θ))=csc^2(θ)

解答

证明

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

解答

真

求解步骤

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

调整左侧

\frac{csc ( θ )}{sin ( θ )}

用 sin, cos 表示

\frac{csc ( θ )}{sin ( θ )}

使用基本三角恒等式:

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

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

化简

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

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

使用分式法则:

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

= \frac{1}{sin ( θ ) sin ( θ )}

sin (θ) sin (θ) = sin^{2} (θ)

sin (θ) sin (θ)

使用指数法则:

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

sin (θ) sin (θ) = sin^{1 + 1} (θ)

= sin^{1 + 1} (θ)

数字相加:

1 + 1 = 2

= sin^{2} (θ)

= \frac{1}{sin ^{2} ( θ )}

= \frac{1}{sin ^{2} ( θ )}

= \frac{1}{sin ^{2} ( θ )}

使用三角恒等式改写

使用基本三角恒等式:

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

\frac{1}{( \frac{1}{c s c ( θ )} ) ^{2}}

化简

\frac{1}{( \frac{1}{c s c ( θ )} ) ^{2}}

(\frac{1}{csc ( θ )})^{2} = \frac{1}{csc ^{2} ( θ )}

(\frac{1}{csc ( θ )})^{2}

使用指数法则:

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

= \frac{1 ^{2}}{csc ^{2} ( θ )}

使用法则

1^{a} = 1

1^{2} = 1

= \frac{1}{csc ^{2} ( θ )}

= \frac{1}{\frac{1}{c s c ^{2} ( θ )}}

使用分式法则:

\frac{1}{\frac{b}{c}} = \frac{c}{b}

= \frac{csc ^{2} ( θ )}{1}

使用法则

\frac{a}{1} = a

= csc^{2} (θ)

csc^{2} (θ)

csc^{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 2sin(2x)=4sin(x)cos(x)

p ro v e 2 sin (2 x) = 4 sin (x) cos (x)

证明 (1-sin(x))/(sec(x))=(cos^3(x))/(1+sin(x))

p ro v e \frac{1 - sin ( x )}{sec ( x )} = \frac{cos ^{3} ( x )}{1 + sin ( x )}

证明 cos(2α)=-1+2/(1+tan^2(α))

p ro v e cos (2 α) = - 1 + \frac{2}{1 + tan ^{2} ( α )}

证明 (1+sec(x))(csc(x)-cot(x))=tan(x)

p ro v e (1 + sec (x)) (csc (x) - cot (x)) = tan (x)

证明 2-csc^2(z)=1-cot^2(z)

p ro v e 2 - csc^{2} (z) = 1 - cot^{2} (z)