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受欢迎的三角函数 >

证明 cos(b)(1+tan(b))^2=sec(b)+2sin(b)

解答

证明

cos (b) (1 + tan (b))^{2} = sec (b) + 2 sin (b)

解答

真

求解步骤

cos (b) (1 + tan (b))^{2} = sec (b) + 2 sin (b)

调整左侧

cos (b) (1 + tan (b))^{2}

用 sin, cos 表示

(1 + tan (b))^{2} cos (b)

使用基本三角恒等式:

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

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

化简

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

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

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

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

化简

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

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

将项转换为分式:

1 = \frac{1 cos ( b )}{cos ( b )}

= \frac{1 \cdot cos ( b )}{cos ( b )} + \frac{sin ( b )}{cos ( 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{1 \cdot cos ( b ) + sin ( b )}{cos ( b )}

乘以:

1 \cdot cos (b) = cos (b)

= \frac{cos ( b ) + sin ( b )}{cos ( b )}

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

使用指数法则:

(\frac{a}{b})^{c} = \frac{a ^{c}}{b ^{c}}

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

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

分式相乘:

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

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

约分:

cos (b)

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

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

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

乘开

(cos (b) + sin (b))^{2} : cos^{2} (b) + 2 cos (b) sin (b) + sin^{2} (b)

(cos (b) + sin (b))^{2}

使用完全平方公式:

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

a = cos (b), b = sin (b)

= cos^{2} (b) + 2 cos (b) sin (b) + sin^{2} (b)

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

使用三角恒等式改写

\frac{cos ^{2} ( b ) + sin ^{2} ( b ) + 2 cos ( b ) sin ( b )}{cos ( b )}

使用毕达哥拉斯恒等式:

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

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

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

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{1 + 2 \cdot \frac{1}{s e c ( b )} sin ( b )}{\frac{1}{s e c ( b )}}

化简

\frac{1 + 2 \cdot \frac{1}{s e c ( b )} sin ( b )}{\frac{1}{s e c ( b )}}

使用分式法则:

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

= \frac{( 1 + 2 \cdot \frac{1}{s e c ( b )} sin ( b ) ) sec ( b )}{1}

2 \cdot \frac{1}{sec ( b )} sin (b) = \frac{2 sin ( b )}{sec ( b )}

2 \cdot \frac{1}{sec ( b )} sin (b)

分式相乘:

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

= \frac{1 \cdot 2 sin ( b )}{sec ( b )}

数字相乘:

1 \cdot 2 = 2

= \frac{2 sin ( b )}{sec ( b )}

= \frac{sec ( b ) ( \frac{2 s i n ( b )}{s e c ( b )} + 1 )}{1}

化简

1 + \frac{2 sin ( b )}{sec ( b )} : \frac{sec ( b ) + 2 sin ( b )}{sec ( b )}

1 + \frac{2 sin ( b )}{sec ( b )}

将项转换为分式:

1 = \frac{1 sec ( b )}{sec ( b )}

= \frac{1 \cdot sec ( b )}{sec ( b )} + \frac{2 sin ( b )}{sec ( 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{1 \cdot sec ( b ) + 2 sin ( b )}{sec ( b )}

乘以:

1 \cdot sec (b) = sec (b)

= \frac{sec ( b ) + 2 sin ( b )}{sec ( b )}

= \frac{\frac{s e c ( b ) + 2 s i n ( b )}{s e c ( b )} sec ( b )}{1}

使用分式法则:

\frac{a}{1} = a

= \frac{sec ( b ) + 2 sin ( b )}{sec ( b )} sec (b)

分式相乘:

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

= \frac{( sec ( b ) + 2 sin ( b ) ) sec ( b )}{sec ( b )}

约分:

sec (b)

= sec (b) + 2 sin (b)

sec (b) + 2 sin (b)

sec (b) + 2 sin (b)

我们已展示，在两侧可以有相同的形式

\Rightarrow 真

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