求证三角恒等式：
$$
\begin{aligned}
k_1 = {{sin(\alpha + \beta)} + 1 \over { cos(\alpha + \beta)} } \text{, } k_2 = {{sin(90^\circ + {\beta \over 2}) - sin\alpha} \over {cos(90^\circ + {\beta \over 2}) - cos\alpha}}
\end{aligned}
$$
$$
\text{证明：对$\beta = 0$时： }
\begin{aligned}
k_1 = { {sin\alpha+1} \over {cos\alpha} }, k_2 = { {1-sin\alpha} \over {-cos\alpha} }
\end{aligned}
$$
$$
\begin{aligned}
\text{左边} & = { {(1+sin\alpha+{{1-sin\alpha} \over {cos\alpha}} cos\alpha)^2 } {{1+k_1^2} \over {(k_1-k_2)^2}} } \\
& = {2^2 } { {1+k_1^2} \over { ({{sin\alpha+1} \over {cos\alpha}} + {{1-sin\alpha} \over {cos\alpha}})^2 }} \\
& = {2^2} { {1+k_1^2} \over ( { {2} \over {cos\alpha}})^2 } \\
& = (1+k_1^2)cos^2\alpha \\
& = [1+ { {(1+sin\alpha)^2} \over {cos^2\alpha}}] cos^2\alpha \\
& = cos^2\alpha + (1+sin\alpha)^2 \\
& = 2+2sin\alpha
\end{aligned}
$$
$$
\begin{aligned}
\text{ 得证。}
\end{aligned}
$$
求证：
$$
\begin{aligned}
(1+sin\alpha - k_2 cos\alpha)^2 {{1+k_1^2} \over {(k_1 - k_2)^2} } & = cos^2\alpha + (1+ sin\alpha)^2 \\
& = 2+2sin\alpha
\end{aligned}
$$
等价于： $$ \begin{aligned} A = (2+2sin\alpha)(sin {\beta \over 2} + cos\alpha + sin\beta + cos(\alpha - {\beta \over 2 }+ \beta))^2 \end{aligned} $$ $$ \begin{aligned} B = (2+2sin(\alpha+\beta))(sin {\beta \over 2} + cos\alpha + cos(\alpha - {\beta \over 2 }))^2 \end{aligned} $$ 证： A = B
用和差化积公式可证。
$$ \begin{aligned} {k_1 - k_2 = } { { sin{\beta \over 2} + cos\alpha + cos(\alpha + \beta - {\beta \over 2}) + sin(\alpha + \beta - \alpha) } \over {(sin{\beta \over 2} + cos\alpha) cos(\alpha + \beta) } } \end{aligned} $$ $$ \begin{aligned} 1+k_1^2 = { {2+2sin(\alpha+\beta)} \over {cos^2(\alpha+\beta)} } \end{aligned} $$ $$ \begin{aligned} {1+k_1^2 \over {(k_1 - k_2)^2} } = (2+2sin(\alpha+\beta)) { { {(sin {\beta \over 2} + cos\alpha)^2} \over {(sin{\beta \over 2}+cos\alpha+ cos(\alpha+{\beta \over 2})+sin\beta)^2} } } \end{aligned} $$ $$ \begin{aligned} (1+sin\alpha - k_2 cos\alpha)^2 = ({ {sin{\beta \over 2} + cos\alpha + cos(\alpha - {\beta \over 2})} \over {sin{\beta \over 2}+ cos\alpha} } )^2 \end{aligned} $$ $$ \begin{aligned} \text{左边} & = {(2+2sin(\alpha+\beta)) } {{ (sin{\beta \over 2} + cos\alpha + cos(\alpha - {\beta \over 2}))^2} \over {(sin{\beta \over 2}+cos\alpha+ cos(\alpha+{\beta \over 2})+sin\beta)^2} } \\ \text{右边} & = 2+2sin\alpha \end{aligned} $$ $$ \begin{aligned} \end{aligned} $$
