zs

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

受欢迎的三角函数 >

证明 sin^2(pi/4-2pi)=cos^2(pi/4-2pi)

解答

证明

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

解答

真

求解步骤

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

调整左侧

sin^{2} (\frac{π}{4} - 2 π)

使用三角恒等式改写

sin (\frac{π}{4} - 2 π)

使用角差恒等式:

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

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

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

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

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

sin (\frac{π}{4}) cos (2 π)

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 (2 π)

cos (2 π) = 1

cos (2 π)

cos (2 π) = cos (0)

cos (2 π)

将

2 π

改写为

2 π + 0

= cos (2 π + 0)

使用周期

cos

:

cos (x + 2 π) = cos (x)

cos (2 π + 0) = cos (0)

= cos (0)

= cos (0)

使用以下普通恒等式:

cos (0) = 1

cos (0)

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}

= 1

= 1

= 1 \cdot \frac{2}{2}

乘以:

\frac{2}{2} \cdot 1 = \frac{2}{2}

= \frac{2}{2}

cos (\frac{π}{4}) sin (2 π) = 0

cos (\frac{π}{4}) sin (2 π)

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} sin (2 π)

sin (2 π) = 0

sin (2 π)

sin (2 π) = sin (0)

sin (2 π)

将

2 π

改写为

2 π + 0

= sin (2 π + 0)

使用周期

sin

:

sin (x + 2 π) = sin (x)

sin (2 π + 0) = sin (0)

= sin (0)

= sin (0)

使用以下普通恒等式:

sin (0) = 0

sin (0)

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}

= 0

= 0

= 0 \cdot \frac{2}{2}

使用法则

0 \cdot a = 0

= 0

= \frac{2}{2} - 0

\frac{2}{2} - 0 = \frac{2}{2}

= \frac{2}{2}

= \frac{2}{2}

= (\frac{2}{2})^{2}

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

(\frac{2}{2})^{2}

使用指数法则:

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

= \frac{( 2 ) ^{2}}{2 ^{2}}

(2)^{2} : 2

使用根式运算法则:

a = a^{\frac{1}{2}}

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

使用指数法则:

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

= 2^{\frac{1}{2} \cdot 2}

\frac{1}{2} \cdot 2 = 1

\frac{1}{2} \cdot 2

分式相乘:

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

= \frac{1 \cdot 2}{2}

约分:

2

= 1

= 2

= \frac{2}{2 ^{2}}

约分:

2

= \frac{1}{2}

= \frac{1}{2}

调整右侧

cos^{2} (\frac{π}{4} - 2 π)

使用三角恒等式改写

cos (\frac{π}{4} - 2 π)

使用角差恒等式:

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

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

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

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

cos (\frac{π}{4}) cos (2 π) = \frac{2}{2}

cos (\frac{π}{4}) cos (2 π)

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 (2 π)

cos (2 π) = 1

cos (2 π)

cos (2 π) = cos (0)

cos (2 π)

将

2 π

改写为

2 π + 0

= cos (2 π + 0)

使用周期

cos

:

cos (x + 2 π) = cos (x)

cos (2 π + 0) = cos (0)

= cos (0)

= cos (0)

使用以下普通恒等式:

cos (0) = 1

cos (0)

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}

= 1

= 1

= 1 \cdot \frac{2}{2}

乘以:

\frac{2}{2} \cdot 1 = \frac{2}{2}

= \frac{2}{2}

sin (\frac{π}{4}) sin (2 π) = 0

sin (\frac{π}{4}) sin (2 π)

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} sin (2 π)

sin (2 π) = 0

sin (2 π)

sin (2 π) = sin (0)

sin (2 π)

将

2 π

改写为

2 π + 0

= sin (2 π + 0)

使用周期

sin

:

sin (x + 2 π) = sin (x)

sin (2 π + 0) = sin (0)

= sin (0)

= sin (0)

使用以下普通恒等式:

sin (0) = 0

sin (0)

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}

= 0

= 0

= 0 \cdot \frac{2}{2}

使用法则

0 \cdot a = 0

= 0

= \frac{2}{2} + 0

\frac{2}{2} + 0 = \frac{2}{2}

= \frac{2}{2}

= \frac{2}{2}

= (\frac{2}{2})^{2}

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

(\frac{2}{2})^{2}

使用指数法则:

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

= \frac{( 2 ) ^{2}}{2 ^{2}}

(2)^{2} : 2

使用根式运算法则:

a = a^{\frac{1}{2}}

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

使用指数法则:

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

= 2^{\frac{1}{2} \cdot 2}

\frac{1}{2} \cdot 2 = 1

\frac{1}{2} \cdot 2

分式相乘:

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

= \frac{1 \cdot 2}{2}

约分:

2

= 1

= 2

= \frac{2}{2 ^{2}}

约分:

2

= \frac{1}{2}

= \frac{1}{2}

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 sec(θ)=tan(θ)csc(θ)

p ro v e sec (θ) = tan (θ) csc (θ)

证明 5tan(2t)+3=3

p ro v e 5 tan (2 t) + 3 = 3

证明 (1+sin(y))(1+sin(-y))=cos^2(y)

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

证明 cos(6x)+cos(2x)=2cos(4x)cos(2x)

p ro v e cos (6 x) + cos (2 x) = 2 cos (4 x) cos (2 x)

证明 cos(x)+cot(x)+sin(x)=csc(x)

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