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受欢迎的三角函数 >

证明 (1-cos(x))(1+cos(x))= 1/(csc^2(x))

解答

证明

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

解答

真

求解步骤

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

调整右侧

\frac{1}{csc ^{2} ( x )}

用 sin, cos 表示

\frac{1}{csc ^{2} ( x )}

使用基本三角恒等式:

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

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

化简

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

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

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

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

使用指数法则:

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

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

使用法则

1^{a} = 1

1^{2} = 1

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

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

使用分式法则:

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

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

使用法则

\frac{a}{1} = a

= sin^{2} (x)

= sin^{2} (x)

= sin^{2} (x)

使用三角恒等式改写

sin^{2} (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)

= 1 - cos^{2} (x)

分解

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

1 - cos^{2} (x)

使用平方差公式:

x^{2} - y^{2} = (x + y) (x - y)

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

= (1 + cos (x)) (1 - cos (x))

= (1 + cos (x)) (1 - cos (x))

= (1 + cos (x)) (1 - cos (x))

= (1 - cos (x)) (1 + 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 csc(x)cos(x)=cot(x)

p ro v e csc (x) cos (x) = cot (x)

证明 sin(pi-x)=sin(x)

p ro v e sin (π - x) = sin (x)

证明 cos(θ)csc(θ)=cot(θ)

p ro v e cos (θ) csc (θ) = cot (θ)

证明 cot(-x)sin(x)=-cos(x)

p ro v e cot (- x) sin (x) = - cos (x)

证明 sin(3pi-x)=sin(x)

p ro v e sin (3 π - x) = sin (x)