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受欢迎的三角函数 >

证明 (cos(A))/(1+sin(A))+(cos(A))/(1-sin(A))=2sec(A)

解答

证明

\frac{cos ( A )}{1 + sin ( A )} + \frac{cos ( A )}{1 - sin ( A )} = 2 sec (A)

解答

真

求解步骤

\frac{cos ( A )}{1 + sin ( A )} + \frac{cos ( A )}{1 - sin ( A )} = 2 sec (A)

调整左侧

\frac{cos ( A )}{1 + sin ( A )} + \frac{cos ( A )}{1 - sin ( A )}

使用三角恒等式改写

\frac{cos ( A )}{1 + sin ( A )} + \frac{cos ( A )}{1 - sin ( A )}

乘以

\frac{1 + sin ( A )}{1 + sin ( A )}

= \frac{cos ( A )}{1 + sin ( A )} + \frac{cos ( A )}{\frac{( 1 - s i n ( A ) ) ( 1 + s i n ( A ) )}{1 + s i n ( A )}}

使用平方差公式:

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

(1 - sin (A)) (1 + sin (A)) = 1 - sin^{2} (A)

= \frac{cos ( A )}{1 + sin ( A )} + \frac{cos ( A )}{\frac{1 - s i n ^{2} ( A )}{1 + s i n ( A )}}

使用毕达哥拉斯恒等式:

1 = cos^{2} (A) + sin^{2} (A)

1 - sin^{2} (A) = cos^{2} (A)

= \frac{cos ( A )}{1 + sin ( A )} + \frac{cos ( A )}{\frac{c o s ^{2} ( A )}{1 + s i n ( A )}}

\frac{cos ( A )}{\frac{c o s ^{2} ( A )}{1 + s i n ( A )}} = \frac{1 + sin ( A )}{cos ( A )}

\frac{cos ( A )}{\frac{c o s ^{2} ( A )}{1 + s i n ( A )}}

使用分式法则:

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

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

约分:

cos (A)

= \frac{1 + sin ( A )}{cos ( A )}

= \frac{cos ( A )}{sin ( A ) + 1} + \frac{sin ( A ) + 1}{cos ( A )}

= \frac{cos ( A )}{sin ( A ) + 1} + \frac{sin ( A ) + 1}{cos ( A )}

化简

\frac{1 + sin ( A )}{cos ( A )} + \frac{cos ( A )}{1 + sin ( A )} : \frac{( 1 + sin ( A ) ) ^{2} + cos ^{2} ( A )}{cos ( A ) ( sin ( A ) + 1 )}

\frac{1 + sin ( A )}{cos ( A )} + \frac{cos ( A )}{1 + sin ( A )}

cos (A), 1 + sin (A)

的最小公倍数:

cos (A) (sin (A) + 1)

cos (A), 1 + sin (A)

最小公倍数 (LCM)

计算出由出现在

cos (A)

或

1 + sin (A)

中的因子组成的表达式

= cos (A) (sin (A) + 1)

根据最小公倍数调整分式

将每个分子乘以其分母转变为最小公倍数所要乘以的同一数值

cos (A) (sin (A) + 1)

对于

\frac{1 + sin ( A )}{cos ( A )} :

将分母和分子乘以

sin (A) + 1

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

对于

\frac{cos ( A )}{1 + sin ( A )} :

将分母和分子乘以

cos (A)

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

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

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 + sin ( A ) ) ^{2} + cos ^{2} ( A )}{cos ( A ) ( sin ( A ) + 1 )}

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

使用三角恒等式改写

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

使用毕达哥拉斯恒等式:

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

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

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

化简

\frac{( 1 + sin ( A ) ) ^{2} + 1 - sin ^{2} ( A )}{( 1 + sin ( A ) ) cos ( A )} : \frac{2}{cos ( A )}

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

乘开

(1 + sin (A))^{2} + 1 - sin^{2} (A) : 2 sin (A) + 2

(1 + sin (A))^{2} + 1 - sin^{2} (A)

(1 + sin (A))^{2} : 1 + 2 sin (A) + sin^{2} (A)

使用完全平方公式:

(a + b)^{2} = a^{2} + 2 ab + b^{2}

a = 1, b = sin (A)

= 1^{2} + 2 \cdot 1 \cdot sin (A) + sin^{2} (A)

化简

1^{2} + 2 \cdot 1 \cdot sin (A) + sin^{2} (A) : 1 + 2 sin (A) + sin^{2} (A)

1^{2} + 2 \cdot 1 \cdot sin (A) + sin^{2} (A)

使用法则

1^{a} = 1

1^{2} = 1

= 1 + 2 \cdot 1 \cdot sin (A) + sin^{2} (A)

数字相乘:

2 \cdot 1 = 2

= 1 + 2 sin (A) + sin^{2} (A)

= 1 + 2 sin (A) + sin^{2} (A)

= 1 + 2 sin (A) + sin^{2} (A) + 1 - sin^{2} (A)

化简

1 + 2 sin (A) + sin^{2} (A) + 1 - sin^{2} (A) : 2 sin (A) + 2

1 + 2 sin (A) + sin^{2} (A) + 1 - sin^{2} (A)

对同类项分组

= 2 sin (A) + sin^{2} (A) - sin^{2} (A) + 1 + 1

同类项相加:

sin^{2} (A) - sin^{2} (A) = 0

= 2 sin (A) + 1 + 1

数字相加:

1 + 1 = 2

= 2 sin (A) + 2

= 2 sin (A) + 2

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

分解

2 sin (A) + 2 : 2 (sin (A) + 1)

2 sin (A) + 2

改写为

= 2 sin (A) + 2 \cdot 1

因式分解出通项

2

= 2 (sin (A) + 1)

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

约分:

sin (A) + 1

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

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

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

使用三角恒等式改写

使用基本三角恒等式:

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

\frac{2}{\frac{1}{s e c ( A )}}

化简

\frac{2}{\frac{1}{s e c ( A )}}

使用分式法则:

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

= \frac{2 sec ( A )}{1}

使用法则

\frac{a}{1} = a

= 2 sec (A)

2 sec (A)

2 sec (A)

我们已展示，在两侧可以有相同的形式

\Rightarrow 真

流行的例子

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

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p ro v e - tan (y) + sec (y) = \frac{cos ( y )}{1 + sin ( y )}

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证明 1/(tan(b))+tan(b)=sec(b)csc(b)

p ro v e \frac{1}{tan ( b )} + tan (b) = sec (b) csc (b)