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受欢迎的三角函数 >

证明 sec(pi/4+a)sec(pi/4-a)=2sec(2a)

解答

证明

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

解答

真

求解步骤

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

调整左侧

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

使用三角恒等式改写

sec (\frac{π}{4} - a)

使用基本三角恒等式:

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

= \frac{1}{cos ( \frac{π}{4} - a )}

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 ( a ) + sin ( \frac{π}{4} ) sin ( a )}

化简

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

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

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

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

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 (a) + sin (\frac{π}{4}) sin (a)

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

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

乘

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

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

分式相乘:

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

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

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

乘

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

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

分式相乘:

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

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

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

合并分式

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

使用法则

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

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

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

使用分式法则:

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

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

因式分解出通项

2

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

消掉

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

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

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 ( a ) + sin ( a ) )}

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 ( a ) + sin ( a )}

数字相减:

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

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

使用根式运算法则:

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

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

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

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

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

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

使用三角恒等式改写

sec (\frac{π}{4} + a)

使用基本三角恒等式:

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

= \frac{1}{cos ( \frac{π}{4} + a )}

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 ( a ) - sin ( \frac{π}{4} ) sin ( a )}

化简

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

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

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

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

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 (a) - sin (\frac{π}{4}) sin (a)

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

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

乘

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

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

分式相乘:

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

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

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

乘

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

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

分式相乘:

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

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

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

合并分式

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

使用法则

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

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

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

使用分式法则:

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

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

因式分解出通项

2

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

消掉

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

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

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 ( a ) - sin ( a ) )}

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 ( a ) - sin ( a )}

数字相减:

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

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

使用根式运算法则:

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

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

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

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

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

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

化简

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

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

分式相乘:

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

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

22 = 2

22

使用根式运算法则:

a a = a

22 = 2

= 2

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

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

调整右侧

2 sec (2 a)

用 sin, cos 表示

2 sec (2 a)

使用基本三角恒等式:

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

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

化简

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

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

分式相乘:

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

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

数字相乘:

1 \cdot 2 = 2

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

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

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

使用三角恒等式改写

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

使用倍角公式:

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

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

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

分解

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

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

使用平方差公式:

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

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

= (cos (a) + sin (a)) (cos (a) - sin (a))

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

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 cot^2(v/2)=((sec(v)+1))/((sec(v)-1))

p ro v e cot^{2} (\frac{v}{2}) = \frac{( sec ( v ) + 1 )}{( sec ( v ) - 1 )}

证明 cot(3)=(cos(3))/(sin(3))

p ro v e cot (3) = \frac{cos ( 3 )}{sin ( 3 )}

证明 cos(2θ)-cos(θ)+1=0

p ro v e cos (2 θ) - cos (θ) + 1 = 0

证明 (-sin(-a))/(cos(-a))=tan(a)

p ro v e \frac{- sin ( - a )}{cos ( - a )} = tan (a)

证明 (csc(z))/(sec(z))=cot(z)

p ro v e \frac{csc ( z )}{sec ( z )} = cot (z)