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受欢迎的三角函数 >

证明 cos(75)=sqrt((1+cos(150))/2)

解答

证明

cos (7 5^{\circ}) = \frac{1 + cos ( 15 0 ^{\circ} )}{2}

解答

真

求解步骤

cos (7 5^{\circ}) = \frac{1 + cos ( 15 0 ^{\circ} )}{2}

调整左侧

cos (7 5^{\circ})

使用三角恒等式改写

cos (7 5^{\circ})

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

使用半角公式:

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

使用倍角公式

cos (2 θ) = 2 cos^{2} (θ) - 1

用

\frac{θ}{2}

替代

θ

cos (θ) = 2 cos^{2} (\frac{θ}{2}) - 1

交换两边

2 cos^{2} (\frac{θ}{2}) = 1 + cos (θ)

两边除以

2

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

Square root both sides

Choose the root sign according to the quadrant of

\frac{θ}{2}

:

range [0, 9 0^{\circ}] [9 0^{\circ}, 18 0^{\circ}] [18 0^{\circ}, 27 0^{\circ}] [27 0^{\circ}, 36 0^{\circ}] quadrant I II III I V sin positive positive negative negative cos positive negative negative positive

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

\frac{1 + cos ( 15 0 ^{\circ} )}{2} = \frac{2 - 3}{2}

\frac{1 + cos ( 15 0 ^{\circ} )}{2}

\frac{1 + cos ( 15 0 ^{\circ} )}{2} = \frac{2 - 3}{4}

\frac{1 + cos ( 15 0 ^{\circ} )}{2}

1 + cos (15 0^{\circ}) = 1 - \frac{3}{2}

1 + cos (15 0^{\circ})

使用以下普通恒等式:

cos (15 0^{\circ}) = - \frac{3}{2}

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 - \frac{3}{2}

= \frac{1 - \frac{3}{2}}{2}

化简

1 - \frac{3}{2} : \frac{2 - 3}{2}

1 - \frac{3}{2}

将项转换为分式:

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

= \frac{1 \cdot 2}{2} - \frac{3}{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 - 3}{2}

数字相乘:

1 \cdot 2 = 2

= \frac{2 - 3}{2}

= \frac{\frac{2 - 3}{2}}{2}

使用分式法则:

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

= \frac{2 - 3}{2 \cdot 2}

数字相乘:

2 \cdot 2 = 4

= \frac{2 - 3}{4}

= \frac{2 - 3}{4}

化简

\frac{2 - 3}{4}

使用根式运算法则:

n \frac{a}{b} = \frac{n a}{n b},

假定

a \geq 0, b \geq 0

= \frac{2 - 3}{4}

4 = 2

4

因式分解数字:

4 = 2^{2}

= 2^{2}

使用根式运算法则:

n a^{n} = a

2^{2} = 2

= 2

= \frac{2 - 3}{2}

= \frac{2 - 3}{2}

= \frac{2 - 3}{2}

= \frac{2 - 3}{2}

调整右侧

\frac{1 + cos ( 15 0 ^{\circ} )}{2}

化简

\frac{1 + cos ( 15 0 ^{\circ} )}{2} : \frac{2 - 3}{2}

\frac{1 + cos ( 15 0 ^{\circ} )}{2}

\frac{1 + cos ( 15 0 ^{\circ} )}{2} = \frac{2 - 3}{4}

\frac{1 + cos ( 15 0 ^{\circ} )}{2}

1 + cos (15 0^{\circ}) = 1 - \frac{3}{2}

1 + cos (15 0^{\circ})

使用以下普通恒等式:

cos (15 0^{\circ}) = - \frac{3}{2}

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 - \frac{3}{2}

= \frac{1 - \frac{3}{2}}{2}

化简

1 - \frac{3}{2} : \frac{2 - 3}{2}

1 - \frac{3}{2}

将项转换为分式:

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

= \frac{1 \cdot 2}{2} - \frac{3}{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 - 3}{2}

数字相乘:

1 \cdot 2 = 2

= \frac{2 - 3}{2}

= \frac{\frac{2 - 3}{2}}{2}

使用分式法则:

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

= \frac{2 - 3}{2 \cdot 2}

数字相乘:

2 \cdot 2 = 4

= \frac{2 - 3}{4}

= \frac{2 - 3}{4}

化简

\frac{2 - 3}{4}

使用根式运算法则:

n \frac{a}{b} = \frac{n a}{n b},

假定

a \geq 0, b \geq 0

= \frac{2 - 3}{4}

4 = 2

4

因式分解数字:

4 = 2^{2}

= 2^{2}

使用根式运算法则:

n a^{n} = a

2^{2} = 2

= 2

= \frac{2 - 3}{2}

= \frac{2 - 3}{2}

= \frac{2 - 3}{2}

我们已展示，在两侧可以有相同的形式

\Rightarrow 真

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