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受欢迎的三角函数 >

证明 tan(pi/4-θ)=(cos(2θ))/(1+sin(2θ))

解答

证明

tan (\frac{π}{4} - θ) = \frac{cos ( 2 θ )}{1 + sin ( 2 θ )}

解答

真

求解步骤

tan (\frac{π}{4} - θ) = \frac{cos ( 2 θ )}{1 + sin ( 2 θ )}

调整左侧

tan (\frac{π}{4} - θ)

使用三角恒等式改写

tan (\frac{π}{4} - θ)

使用基本三角恒等式:

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

= \frac{sin ( \frac{π}{4} - θ )}{cos ( \frac{π}{4} - θ )}

使用角差恒等式:

sin (s - t) = sin (s) cos (t) - cos (s) sin (t)

= \frac{sin ( \frac{π}{4} ) cos ( θ ) - cos ( \frac{π}{4} ) sin ( θ )}{cos ( \frac{π}{4} - θ )}

使用角差恒等式:

cos (s - t) = cos (s) cos (t) + sin (s) sin (t)

= \frac{sin ( \frac{π}{4} ) cos ( θ ) - cos ( \frac{π}{4} ) sin ( θ )}{cos ( \frac{π}{4} ) cos ( θ ) + sin ( \frac{π}{4} ) sin ( θ )}

化简

\frac{sin ( \frac{π}{4} ) cos ( θ ) - cos ( \frac{π}{4} ) sin ( θ )}{cos ( \frac{π}{4} ) cos ( θ ) + sin ( \frac{π}{4} ) sin ( θ )} : \frac{cos ( θ ) - sin ( θ )}{cos ( θ ) + sin ( θ )}

\frac{sin ( \frac{π}{4} ) cos ( θ ) - cos ( \frac{π}{4} ) sin ( θ )}{cos ( \frac{π}{4} ) cos ( θ ) + sin ( \frac{π}{4} ) sin ( θ )}

sin (\frac{π}{4}) cos (θ) - cos (\frac{π}{4}) sin (θ) = \frac{2}{2} cos (θ) - \frac{2}{2} sin (θ)

sin (\frac{π}{4}) cos (θ) - cos (\frac{π}{4}) sin (θ)

化简

sin (\frac{π}{4}) : \frac{2}{2}

sin (\frac{π}{4})

使用以下普通恒等式:

sin (\frac{π}{4}) = \frac{2}{2}

sin (x)

周期表（周期为

2 πn

"）：

x 0 \frac{π}{6} \frac{π}{4} \frac{π}{3} \frac{π}{2} \frac{2 π}{3} \frac{3 π}{4} \frac{5 π}{6} sin (x) 0 \frac{1}{2} \frac{2}{2} \frac{3}{2} 1 \frac{3}{2} \frac{2}{2} \frac{1}{2} x π \frac{7 π}{6} \frac{5 π}{4} \frac{4 π}{3} \frac{3 π}{2} \frac{5 π}{3} \frac{7 π}{4} \frac{11 π}{6} sin (x) 0 - \frac{1}{2} - \frac{2}{2} - \frac{3}{2} - 1 - \frac{3}{2} - \frac{2}{2} - \frac{1}{2}

= \frac{2}{2}

= \frac{2}{2} cos (θ) - cos (\frac{π}{4}) sin (θ)

化简

cos (\frac{π}{4}) : \frac{2}{2}

cos (\frac{π}{4})

使用以下普通恒等式:

cos (\frac{π}{4}) = \frac{2}{2}

cos (x)

周期表（周期为

2 πn

）：

x 0 \frac{π}{6} \frac{π}{4} \frac{π}{3} \frac{π}{2} \frac{2 π}{3} \frac{3 π}{4} \frac{5 π}{6} cos (x) 1 \frac{3}{2} \frac{2}{2} \frac{1}{2} 0 - \frac{1}{2} - \frac{2}{2} - \frac{3}{2} x π \frac{7 π}{6} \frac{5 π}{4} \frac{4 π}{3} \frac{3 π}{2} \frac{5 π}{3} \frac{7 π}{4} \frac{11 π}{6} cos (x) - 1 - \frac{3}{2} - \frac{2}{2} - \frac{1}{2} 0 \frac{1}{2} \frac{2}{2} \frac{3}{2}

= \frac{2}{2}

= \frac{2}{2} cos (θ) - \frac{2}{2} sin (θ)

= \frac{\frac{2}{2} cos ( θ ) - \frac{2}{2} sin ( θ )}{cos ( \frac{π}{4} ) cos ( θ ) + sin ( \frac{π}{4} ) sin ( θ )}

cos (\frac{π}{4}) cos (θ) + sin (\frac{π}{4}) sin (θ) = \frac{2}{2} cos (θ) + \frac{2}{2} sin (θ)

cos (\frac{π}{4}) cos (θ) + sin (\frac{π}{4}) sin (θ)

化简

cos (\frac{π}{4}) : \frac{2}{2}

cos (\frac{π}{4})

使用以下普通恒等式:

cos (\frac{π}{4}) = \frac{2}{2}

cos (x)

周期表（周期为

2 πn

）：

x 0 \frac{π}{6} \frac{π}{4} \frac{π}{3} \frac{π}{2} \frac{2 π}{3} \frac{3 π}{4} \frac{5 π}{6} cos (x) 1 \frac{3}{2} \frac{2}{2} \frac{1}{2} 0 - \frac{1}{2} - \frac{2}{2} - \frac{3}{2} x π \frac{7 π}{6} \frac{5 π}{4} \frac{4 π}{3} \frac{3 π}{2} \frac{5 π}{3} \frac{7 π}{4} \frac{11 π}{6} cos (x) - 1 - \frac{3}{2} - \frac{2}{2} - \frac{1}{2} 0 \frac{1}{2} \frac{2}{2} \frac{3}{2}

= \frac{2}{2}

= \frac{2}{2} cos (θ) + sin (\frac{π}{4}) sin (θ)

化简

sin (\frac{π}{4}) : \frac{2}{2}

sin (\frac{π}{4})

使用以下普通恒等式:

sin (\frac{π}{4}) = \frac{2}{2}

sin (x)

周期表（周期为

2 πn

"）：

x 0 \frac{π}{6} \frac{π}{4} \frac{π}{3} \frac{π}{2} \frac{2 π}{3} \frac{3 π}{4} \frac{5 π}{6} sin (x) 0 \frac{1}{2} \frac{2}{2} \frac{3}{2} 1 \frac{3}{2} \frac{2}{2} \frac{1}{2} x π \frac{7 π}{6} \frac{5 π}{4} \frac{4 π}{3} \frac{3 π}{2} \frac{5 π}{3} \frac{7 π}{4} \frac{11 π}{6} sin (x) 0 - \frac{1}{2} - \frac{2}{2} - \frac{3}{2} - 1 - \frac{3}{2} - \frac{2}{2} - \frac{1}{2}

= \frac{2}{2}

= \frac{2}{2} cos (θ) + \frac{2}{2} sin (θ)

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

乘

\frac{2}{2} cos (θ) : \frac{2 cos ( θ )}{2}

\frac{2}{2} cos (θ)

分式相乘:

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

= \frac{2 cos ( θ )}{2}

= \frac{\frac{2}{2} cos ( θ ) - \frac{2}{2} sin ( θ )}{\frac{2 c o s ( θ )}{2} + \frac{2}{2} sin ( θ )}

乘

\frac{2}{2} sin (θ) : \frac{2 sin ( θ )}{2}

\frac{2}{2} sin (θ)

分式相乘:

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

= \frac{2 sin ( θ )}{2}

= \frac{\frac{2}{2} cos ( θ ) - \frac{2}{2} sin ( θ )}{\frac{2 c o s ( θ )}{2} + \frac{2 s i n ( θ )}{2}}

乘

\frac{2}{2} cos (θ) : \frac{2 cos ( θ )}{2}

\frac{2}{2} cos (θ)

分式相乘:

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

= \frac{2 cos ( θ )}{2}

= \frac{\frac{2 c o s ( θ )}{2} - \frac{2}{2} sin ( θ )}{\frac{2 c o s ( θ )}{2} + \frac{2 s i n ( θ )}{2}}

乘

\frac{2}{2} sin (θ) : \frac{2 sin ( θ )}{2}

\frac{2}{2} sin (θ)

分式相乘:

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

= \frac{2 sin ( θ )}{2}

= \frac{\frac{2 c o s ( θ )}{2} - \frac{2 s i n ( θ )}{2}}{\frac{2 c o s ( θ )}{2} + \frac{2 s i n ( θ )}{2}}

合并分式

\frac{2 cos ( θ )}{2} + \frac{2 sin ( θ )}{2} : \frac{2 cos ( θ ) + 2 sin ( θ )}{2}

使用法则

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

= \frac{2 cos ( θ ) + 2 sin ( θ )}{2}

= \frac{\frac{2 c o s ( θ )}{2} - \frac{2 s i n ( θ )}{2}}{\frac{2 c o s ( θ ) + 2 s i n ( θ )}{2}}

合并分式

\frac{2 cos ( θ )}{2} - \frac{2 sin ( θ )}{2} : \frac{2 cos ( θ ) - 2 sin ( θ )}{2}

使用法则

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

= \frac{2 cos ( θ ) - 2 sin ( θ )}{2}

= \frac{\frac{2 c o s ( θ ) - 2 s i n ( θ )}{2}}{\frac{2 c o s ( θ ) + 2 s i n ( θ )}{2}}

分式相除:

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

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

约分:

2

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

因式分解出通项

2

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

因式分解出通项

2

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

约分:

2

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

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

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

乘以

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

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

乘开

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

(cos (θ) - sin (θ)) (cos (θ) - sin (θ)) cos (2 θ)

使用指数法则:

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

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

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

数字相加:

1 + 1 = 2

= (cos (θ) - sin (θ))^{2} cos (2 θ)

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

(cos (θ) - sin (θ))^{2}

使用完全平方公式:

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

a = cos (θ), b = sin (θ)

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

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

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

打开括号

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

使用加减运算法则

+ (- a) = - a

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

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

使用三角恒等式改写

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

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

化简

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

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

分解

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

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

因式分解出通项

cos (2 θ)

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

整理后得

= cos (2 θ) (- 2 cos (θ) sin (θ) + 1)

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

约分:

cos (2 θ)

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

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

使用倍角公式:

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

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

乘开

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

(cos (θ) + sin (θ)) (cos (θ) - sin (θ))

使用平方差公式:

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

a = cos (θ), b = sin (θ)

= cos^{2} (θ) - sin^{2} (θ)

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

使用倍角公式:

cos^{2} (θ) - sin^{2} (θ) = cos (2 θ)

= \frac{1 - sin ( 2 θ )}{cos ( 2 θ )}

乘以

\frac{1 + sin ( 2 θ )}{1 + sin ( 2 θ )}

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

使用平方差公式:

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

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

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

使用毕达哥拉斯恒等式:

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

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

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

约分:

cos (2 θ)

= \frac{cos ( 2 θ )}{1 + sin ( 2 θ )}

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

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

证明 (csc(α)+cot(α))(sec(α)-1)=tan(α)

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

证明 1-2sin^2(θ)=2cos^2(θ)-1

p ro v e 1 - 2 sin^{2} (θ) = 2 cos^{2} (θ) - 1

证明 sin(pi/4)=cos(pi/4)

p ro v e sin (\frac{π}{4}) = cos (\frac{π}{4})

证明 sin(3a)+sin(a)=4sin(a)-4sin(3a)

p ro v e sin (3 a) + sin (a) = 4 sin (a) - 4 sin (3 a)