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受欢迎的三角函数 >

证明 (1-tan(x))/(1+tan(x))=(1-sin(2x))/(cos(2x))

解答

证明

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

解答

真

求解步骤

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

调整左侧

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

用 sin, cos 表示

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

使用基本三角恒等式:

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

= \frac{1 - \frac{s i n ( x )}{c o s ( x )}}{1 + \frac{s i n ( x )}{c o s ( x )}}

化简

\frac{1 - \frac{s i n ( x )}{c o s ( x )}}{1 + \frac{s i n ( x )}{c o s ( x )}} : \frac{cos ( x ) - sin ( x )}{cos ( x ) + sin ( x )}

\frac{1 - \frac{s i n ( x )}{c o s ( x )}}{1 + \frac{s i n ( x )}{c o s ( x )}}

化简

1 + \frac{sin ( x )}{cos ( x )} : \frac{cos ( x ) + sin ( x )}{cos ( x )}

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

将项转换为分式:

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

= \frac{1 \cdot cos ( x )}{cos ( x )} + \frac{sin ( x )}{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{1 \cdot cos ( x ) + sin ( x )}{cos ( x )}

乘以:

1 \cdot cos (x) = cos (x)

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

= \frac{1 - \frac{s i n ( x )}{c o s ( x )}}{\frac{c o s ( x ) + s i n ( x )}{c o s ( x )}}

化简

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

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

将项转换为分式:

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

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

乘以:

1 \cdot cos (x) = cos (x)

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

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

分式相除:

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

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

约分:

cos (x)

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

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

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

乘以

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

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

乘开

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

(cos (x) - sin (x)) (cos (x) - sin (x)) cos (2 x)

使用指数法则:

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

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

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

数字相加:

1 + 1 = 2

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

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

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

使用完全平方公式:

(a - b)^{2} = a^{2} - 2 ab + b^{2}

a = cos (x), b = sin (x)

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

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

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

打开括号

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

使用加减运算法则

+ (- a) = - a

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

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

使用三角恒等式改写

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

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

化简

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

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

分解

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

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

因式分解出通项

cos (2 x)

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

整理后得

= cos (2 x) (- 2 cos (x) sin (x) + 1)

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

约分:

cos (2 x)

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

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

使用倍角公式:

2 sin (x) cos (x) = sin (2 x)

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

乘开

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

(cos (x) + sin (x)) (cos (x) - sin (x))

使用平方差公式:

(a + b) (a - b) = a^{2} - b^{2}

a = cos (x), b = sin (x)

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

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

使用倍角公式:

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

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

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

p ro v e csch^{2} (x) = coth^{2} (x) - 1

证明 (tan(y))/(csc(y))= 1/(cos(y))-1/(sec(y))

p ro v e \frac{tan ( y )}{csc ( y )} = \frac{1}{cos ( y )} - \frac{1}{sec ( y )}

证明 2tan(x)sec(x)= 1/(1-sin(x))-1/(1+sin(x))

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

证明 1/(sin(x)cot(x))= 1/(cos(x))

p ro v e \frac{1}{sin ( x ) cot ( x )} = \frac{1}{cos ( x )}

证明 (csc^2(x)-1)/(csc^2(x))=cos^2(x)

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