2019河南模考题

 

 

$$\cos \left( A-B\right) =\frac {2\sin A\sin B} {\sin C}
\Rightarrow \sin C \cdot \cos \left( A-B\right) =\cos \left( A-B\right) -\cos \left( A+B\right) ,$$

$$\cos \left( A+B \right) =\left( 1-\sin C\right) \cdot \cos\left( A-B\right) \Rightarrow -\cos C=\left( 1-\sin C\right) \cdot \cos \left( A-B \right) $$

$$\Rightarrow \sin ^{2}\frac {C} {2}-\cos ^{2}\frac {C} {2} =\left( \sin ^{2}\frac {C} {2}-2\cdot \sin \frac {C} {2}\cos \frac {C} {2}+\cos ^{2}\frac {C} {2}\right) \cdot \cos \left( A-B \right) $$
$$\Rightarrow \sin ^{2}\frac {C} {2}-\cos ^{2}\frac {C} {2} =\left( \sin \frac {C} {2}-\cos \frac {C} {2}\right) ^{2}\cdot \cos \left( A-B\right) $$

$\frac { \sin ^{2}\frac {C} {2}-\cos ^{2}\frac {C} {2} } {\left( \sin \frac {C} {2}-\cos \frac {C} {2}\right) ^{2}}=\cos \left( A-B\right) \Rightarrow \frac {\sin \frac {C} {2}+\cos\frac {C} {2}} {\sin \frac {C} {2}-\cos \frac {C} {2}}=\cos\left( A-B\right) $

$\frac {\sin \frac {C} {2}+\cos \frac {C} {2}} {\left| \sin \frac {C} {2}-\cos \frac {C} {2} \right| }>1$

$C_{\triangle ABC}=a+b+\sqrt {a^{2}+b^{2}}$

$\left( a^{2}+b^{2}\right) \left( 4^{2}+3^{2}\right) \geq \left( 4a+3b\right) ^{2}$

$a+b+\sqrt {a^{2}+b^{2}}\geq a+b+\frac {4a+3b} {5}-\frac {9} {5}a+\frac {8} {5}b$

$\left( \frac {8} {5}a+\frac {9} {5}b\right) \cdot 1-\left( \frac {9} {5}a+\frac {8} {5}b\right) \cdot \left( \frac {2} {a}+\frac {1} {b}\right) =\frac {2b} {5}+\frac {1b} {5}\cdot \frac {b} {a}+\frac {9} {5}\cdot \frac {b} {a}\geq \frac {2b} {5}+2\cdot \sqrt {\frac {16} {5}\times \frac {9} {5}\times \frac {b} {a}\times \frac {a} {b}}=\frac {26} {5}+\frac {24} {5}=10$