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 cos(300)=1-2sin^2(150)

解答

证明

cos (30 0^{\circ}) = 1 - 2 sin^{2} (15 0^{\circ})

解答

真

求解步骤

cos (30 0^{\circ}) = 1 - 2 sin^{2} (15 0^{\circ})

调整左侧

cos (30 0^{\circ})

化简

cos (30 0^{\circ}) : \frac{1}{2}

cos (30 0^{\circ})

使用三角恒等式改写:

cos (18 0^{\circ}) cos (12 0^{\circ}) - sin (18 0^{\circ}) sin (12 0^{\circ})

cos (30 0^{\circ})

将

cos (30 0^{\circ})

写为

cos (18 0^{\circ} + 12 0^{\circ})

= cos (18 0^{\circ} + 12 0^{\circ})

使用角和恒等式:

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

= cos (18 0^{\circ}) cos (12 0^{\circ}) - sin (18 0^{\circ}) sin (12 0^{\circ})

= cos (18 0^{\circ}) cos (12 0^{\circ}) - sin (18 0^{\circ}) sin (12 0^{\circ})

使用以下普通恒等式:

cos (18 0^{\circ}) = (- 1)

cos (18 0^{\circ})

cos (x)

周期表（周期为

36 0^{\circ} n

）：

x 0 3 0^{\circ} 4 5^{\circ} 6 0^{\circ} 9 0^{\circ} 12 0^{\circ} 13 5^{\circ} 15 0^{\circ} cos (x) 1 \frac{3}{2} \frac{2}{2} \frac{1}{2} 0 - \frac{1}{2} - \frac{2}{2} - \frac{3}{2} x 18 0^{\circ} 21 0^{\circ} 22 5^{\circ} 24 0^{\circ} 27 0^{\circ} 30 0^{\circ} 31 5^{\circ} 33 0^{\circ} cos (x) - 1 - \frac{3}{2} - \frac{2}{2} - \frac{1}{2} 0 \frac{1}{2} \frac{2}{2} \frac{3}{2}

= (- 1)

使用以下普通恒等式:

cos (12 0^{\circ}) = - \frac{1}{2}

cos (12 0^{\circ})

cos (x)

周期表（周期为

36 0^{\circ} n

）：

x 0 3 0^{\circ} 4 5^{\circ} 6 0^{\circ} 9 0^{\circ} 12 0^{\circ} 13 5^{\circ} 15 0^{\circ} cos (x) 1 \frac{3}{2} \frac{2}{2} \frac{1}{2} 0 - \frac{1}{2} - \frac{2}{2} - \frac{3}{2} x 18 0^{\circ} 21 0^{\circ} 22 5^{\circ} 24 0^{\circ} 27 0^{\circ} 30 0^{\circ} 31 5^{\circ} 33 0^{\circ} cos (x) - 1 - \frac{3}{2} - \frac{2}{2} - \frac{1}{2} 0 \frac{1}{2} \frac{2}{2} \frac{3}{2}

= - \frac{1}{2}

使用以下普通恒等式:

sin (18 0^{\circ}) = 0

sin (18 0^{\circ})

sin (x)

周期表（周期为

36 0^{\circ} n

"）：

x 0 3 0^{\circ} 4 5^{\circ} 6 0^{\circ} 9 0^{\circ} 12 0^{\circ} 13 5^{\circ} 15 0^{\circ} sin (x) 0 \frac{1}{2} \frac{2}{2} \frac{3}{2} 1 \frac{3}{2} \frac{2}{2} \frac{1}{2} x 18 0^{\circ} 21 0^{\circ} 22 5^{\circ} 24 0^{\circ} 27 0^{\circ} 30 0^{\circ} 31 5^{\circ} 33 0^{\circ} sin (x) 0 - \frac{1}{2} - \frac{2}{2} - \frac{3}{2} - 1 - \frac{3}{2} - \frac{2}{2} - \frac{1}{2}

= 0

使用以下普通恒等式:

sin (12 0^{\circ}) = \frac{3}{2}

sin (12 0^{\circ})

sin (x)

周期表（周期为

36 0^{\circ} n

"）：

x 0 3 0^{\circ} 4 5^{\circ} 6 0^{\circ} 9 0^{\circ} 12 0^{\circ} 13 5^{\circ} 15 0^{\circ} sin (x) 0 \frac{1}{2} \frac{2}{2} \frac{3}{2} 1 \frac{3}{2} \frac{2}{2} \frac{1}{2} x 18 0^{\circ} 21 0^{\circ} 22 5^{\circ} 24 0^{\circ} 27 0^{\circ} 30 0^{\circ} 31 5^{\circ} 33 0^{\circ} sin (x) 0 - \frac{1}{2} - \frac{2}{2} - \frac{3}{2} - 1 - \frac{3}{2} - \frac{2}{2} - \frac{1}{2}

= \frac{3}{2}

= (- 1) (- \frac{1}{2}) - 0 \cdot \frac{3}{2}

化简

= \frac{1}{2}

= \frac{1}{2}

调整右侧

1 - 2 sin^{2} (15 0^{\circ})

化简

1 - 2 sin^{2} (15 0^{\circ}) : \frac{1}{2}

1 - 2 sin^{2} (15 0^{\circ})

2 sin^{2} (15 0^{\circ}) = \frac{1}{2}

2 sin^{2} (15 0^{\circ})

sin^{2} (15 0^{\circ}) = \frac{1}{2 ^{2}}

sin^{2} (15 0^{\circ})

化简

sin (15 0^{\circ}) : \frac{1}{2}

sin (15 0^{\circ})

使用以下普通恒等式:

sin (15 0^{\circ}) = \frac{1}{2}

sin (x)

周期表（周期为

36 0^{\circ} n

"）：

x 0 3 0^{\circ} 4 5^{\circ} 6 0^{\circ} 9 0^{\circ} 12 0^{\circ} 13 5^{\circ} 15 0^{\circ} sin (x) 0 \frac{1}{2} \frac{2}{2} \frac{3}{2} 1 \frac{3}{2} \frac{2}{2} \frac{1}{2} x 18 0^{\circ} 21 0^{\circ} 22 5^{\circ} 24 0^{\circ} 27 0^{\circ} 30 0^{\circ} 31 5^{\circ} 33 0^{\circ} sin (x) 0 - \frac{1}{2} - \frac{2}{2} - \frac{3}{2} - 1 - \frac{3}{2} - \frac{2}{2} - \frac{1}{2}

= \frac{1}{2}

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

使用指数法则:

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

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

使用法则

1^{a} = 1

1^{2} = 1

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

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

分式相乘:

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

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

约分:

2

= \frac{1}{2}

= 1 - \frac{1}{2}

化简

1 - \frac{1}{2}

将项转换为分式:

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

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

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 2 - 1}{2}

1 \cdot 2 - 1 = 1

1 \cdot 2 - 1

数字相乘:

1 \cdot 2 = 2

= 2 - 1

数字相减:

2 - 1 = 1

= 1

= \frac{1}{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 csc^2(x)+3cot^2(x)-5=4(cot(x)-1)

p ro v e csc^{2} (x) + 3 cot^{2} (x) - 5 = 4 (cot (x) - 1)

证明 (3)((cos(2z))^2)/2 =(3cos(4z))/4

p ro v e (3) \frac{( cos ( 2 z ) ) ^{2}}{2} = \frac{3 cos ( 4 z )}{4}

证明-2sin^2(x)+cos(x)+1=0

p ro v e - 2 sin^{2} (x) + cos (x) + 1 = 0

证明 (2cot(u))/(csc^2(u)-2)=tan(2u)

p ro v e \frac{2 cot ( u )}{csc ^{2} ( u ) - 2} = tan (2 u)

证明 csc(2x)+cot(2x)=(1+cos(2x))/(sin(2x))

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