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受欢迎的三角函数 >

证明 sec(pi/4+x)sec(pi/4-x)=2sec(2x)

解答

证明

sec (\frac{π}{4} + x) sec (\frac{π}{4} - x) = 2 sec (2 x)

解答

真

求解步骤

sec (\frac{π}{4} + x) sec (\frac{π}{4} - x) = 2 sec (2 x)

调整左侧

sec (\frac{π}{4} + x) sec (\frac{π}{4} - x)

使用三角恒等式改写

sec (\frac{π}{4} - x)

使用基本三角恒等式:

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

= \frac{1}{cos ( \frac{π}{4} - x )}

使用角差恒等式:

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

= \frac{1}{cos ( \frac{π}{4} ) cos ( x ) + sin ( \frac{π}{4} ) sin ( x )}

化简

\frac{1}{cos ( \frac{π}{4} ) cos ( x ) + sin ( \frac{π}{4} ) sin ( x )} : \frac{2}{cos ( x ) + sin ( x )}

\frac{1}{cos ( \frac{π}{4} ) cos ( x ) + sin ( \frac{π}{4} ) sin ( x )}

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

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

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 (x) + sin (\frac{π}{4}) sin (x)

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 (x) + \frac{2}{2} sin (x)

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

乘

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

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

分式相乘:

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

= \frac{2 cos ( x )}{2}

= \frac{1}{\frac{2 c o s ( x )}{2} + \frac{2}{2} sin ( x )}

乘

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

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

分式相乘:

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

= \frac{2 sin ( x )}{2}

= \frac{1}{\frac{2 c o s ( x )}{2} + \frac{2 s i n ( x )}{2}}

合并分式

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

使用法则

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

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

= \frac{1}{\frac{2 c o s ( x ) + 2 s i n ( x )}{2}}

使用分式法则:

\frac{1}{\frac{b}{c}} = \frac{c}{b}

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

因式分解出通项

2

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

消掉

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

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

使用根式运算法则:

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

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

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

使用指数法则:

\frac{x ^{a}}{x ^{b}} = x^{a - b}

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

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

数字相减:

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

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

使用根式运算法则:

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

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

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

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

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

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

使用三角恒等式改写

sec (\frac{π}{4} + x)

使用基本三角恒等式:

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

= \frac{1}{cos ( \frac{π}{4} + x )}

使用角和恒等式:

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

= \frac{1}{cos ( \frac{π}{4} ) cos ( x ) - sin ( \frac{π}{4} ) sin ( x )}

化简

\frac{1}{cos ( \frac{π}{4} ) cos ( x ) - sin ( \frac{π}{4} ) sin ( x )} : \frac{2}{cos ( x ) - sin ( x )}

\frac{1}{cos ( \frac{π}{4} ) cos ( x ) - sin ( \frac{π}{4} ) sin ( x )}

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

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

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 (x) - sin (\frac{π}{4}) sin (x)

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

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

乘

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

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

分式相乘:

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

= \frac{2 cos ( x )}{2}

= \frac{1}{\frac{2 c o s ( x )}{2} - \frac{2}{2} sin ( x )}

乘

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

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

分式相乘:

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

= \frac{2 sin ( x )}{2}

= \frac{1}{\frac{2 c o s ( x )}{2} - \frac{2 s i n ( x )}{2}}

合并分式

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

使用法则

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

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

= \frac{1}{\frac{2 c o s ( x ) - 2 s i n ( x )}{2}}

使用分式法则:

\frac{1}{\frac{b}{c}} = \frac{c}{b}

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

因式分解出通项

2

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

消掉

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

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

使用根式运算法则:

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

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

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

使用指数法则:

\frac{x ^{a}}{x ^{b}} = x^{a - b}

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

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

数字相减:

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

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

使用根式运算法则:

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

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

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

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

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

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

化简

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

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

分式相乘:

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

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

22 = 2

22

使用根式运算法则:

a a = a

22 = 2

= 2

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

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

调整右侧

2 sec (2 x)

用 sin, cos 表示

2 sec (2 x)

使用基本三角恒等式:

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

= 2 \cdot \frac{1}{cos ( 2 x )}

化简

2 \cdot \frac{1}{cos ( 2 x )} : \frac{2}{cos ( 2 x )}

2 \cdot \frac{1}{cos ( 2 x )}

分式相乘:

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

= \frac{1 \cdot 2}{cos ( 2 x )}

数字相乘:

1 \cdot 2 = 2

= \frac{2}{cos ( 2 x )}

= \frac{2}{cos ( 2 x )}

= \frac{2}{cos ( 2 x )}

使用三角恒等式改写

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

使用倍角公式:

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

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

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

分解

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

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

使用平方差公式:

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

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

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

= \frac{2}{( cos ( x ) + sin ( x ) ) ( cos ( x ) - sin ( 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 2sin^3(x)cos(x)+2sin(x)cos^3(x)=sin(2x)

p ro v e 2 sin^{3} (x) cos (x) + 2 sin (x) cos^{3} (x) = sin (2 x)

证明 tan^2(α)(1-sin^2(α))=sin^2(α)

p ro v e tan^{2} (α) (1 - sin^{2} (α)) = sin^{2} (α)

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p ro v e sin (x) = tan (x) cos (x)

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p ro v e tan (2 x) = 2 sin (x) cos (x) sec (2 x)

证明 tan(x)csc(x)= 1/(cos(x))

p ro v e tan (x) csc (x) = \frac{1}{cos ( x )}