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受欢迎的三角函数 >

证明 tan(A+B)+tan(A-B)=((2sin(2A)))/((cos(2A)+cos(2B)))

解答

证明

tan (A + B) + tan (A - B) = \frac{( 2 sin ( 2 A ) )}{( cos ( 2 A ) + cos ( 2 B ) )}

解答

真

求解步骤

tan (A + B) + tan (A - B) = \frac{2 sin ( 2 A )}{cos ( 2 A ) + cos ( 2 B )}

调整左侧

tan (A + B) + tan (A - B)

用 sin, cos 表示

tan (A + B) + tan (A - B)

使用基本三角恒等式:

tan (x) = \frac{sin ( x )}{cos ( x )}

= \frac{sin ( A + B )}{cos ( A + B )} + tan (A - B)

使用基本三角恒等式:

tan (x) = \frac{sin ( x )}{cos ( x )}

= \frac{sin ( A + B )}{cos ( A + B )} + \frac{sin ( A - B )}{cos ( A - B )}

化简

\frac{sin ( A + B )}{cos ( A + B )} + \frac{sin ( A - B )}{cos ( A - B )} : \frac{sin ( A + B ) cos ( A - B ) + sin ( A - B ) cos ( A + B )}{cos ( A + B ) cos ( A - B )}

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

cos (A + B), cos (A - B)

的最小公倍数:

cos (A + B) cos (A - B)

cos (A + B), cos (A - B)

最小公倍数 (LCM)

计算出由出现在

cos (A + B)

或

cos (A - B)

中的因子组成的表达式

= cos (A + B) cos (A - B)

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 + B) cos (A - B)

对于

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

将分母和分子乘以

cos (A - B)

\frac{sin ( A + B )}{cos ( A + B )} = \frac{sin ( A + B ) cos ( A - B )}{cos ( A + B ) cos ( A - B )}

对于

\frac{sin ( A - B )}{cos ( A - B )} :

将分母和分子乘以

cos (A + B)

\frac{sin ( A - B )}{cos ( A - B )} = \frac{sin ( A - B ) cos ( A + B )}{cos ( A - B ) cos ( A + B )}

= \frac{sin ( A + B ) cos ( A - B )}{cos ( A + B ) cos ( A - B )} + \frac{sin ( A - B ) cos ( A + B )}{cos ( A - B ) cos ( A + B )}

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{sin ( A + B ) cos ( A - B ) + sin ( A - B ) cos ( A + B )}{cos ( A + B ) cos ( A - B )}

= \frac{sin ( A + B ) cos ( A - B ) + sin ( A - B ) cos ( A + B )}{cos ( A + B ) cos ( A - B )}

= \frac{cos ( A + B ) sin ( A - B ) + cos ( A - B ) sin ( A + B )}{cos ( A + B ) cos ( A - B )}

使用三角恒等式改写

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

使用积化和差恒等式:

cos (s) cos (t) = \frac{1}{2} (cos (s - t) + cos (s + t))

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

化简

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

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

化简

A + B + A - B : 2 A

A + B + A - B

对同类项分组

= A + A + B - B

同类项相加:

A + A = 2 A

= 2 A + B - B

同类项相加:

B - B = 0

= 2 A

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

乘

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

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

分式相乘:

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

= \frac{1 \cdot ( cos ( A + B - ( A - B ) ) + cos ( 2 A ) )}{2}

1 \cdot (cos (A + B - (A - B)) + cos (2 A)) = cos (A + B - (A - B)) + cos (2 A)

1 \cdot (cos (A + B - (A - B)) + cos (2 A))

乘以:

1 \cdot (cos (A + B - (A - B)) + cos (2 A)) = (cos (A + B - (A - B)) + cos (2 A))

= (cos (A + B - (A - B)) + cos (2 A))

去除括号:

(a) = a

= cos (A + B - (A - B)) + cos (2 A)

= \frac{cos ( A + B - ( A - B ) ) + cos ( 2 A )}{2}

乘开

A + B - (A - B) : 2 B

A + B - (A - B)

- (A - B) : - A + B

- (A - B)

打开括号

= - (A) - (- B)

使用加减运算法则

- (- a) = a, - (a) = - a

= - A + B

= A + B - A + B

化简

A + B - A + B : 2 B

A + B - A + B

对同类项分组

= A - A + B + B

同类项相加:

A - A = 0

= B + B

同类项相加:

B + B = 2 B

= 2 B

= 2 B

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

= \frac{cos ( A + B ) sin ( A - B ) + cos ( A - B ) sin ( A + B )}{\frac{c o s ( 2 B ) + c o s ( 2 A )}{2}}

使用分式法则:

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

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

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

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

改写为

= sin (A + B) cos (A - B) + cos (A + B) sin (A - B)

使用三角恒等式改写

sin (A + B) cos (A - B) + cos (A + B) sin (A - B)

使用角和恒等式:

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

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

化简

A + B + A - B : 2 A

A + B + A - B

对同类项分组

= A + A + B - B

同类项相加:

A + A = 2 A

= 2 A + B - B

同类项相加:

B - B = 0

= 2 A

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

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

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

我们已展示，在两侧可以有相同的形式

\Rightarrow 真

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