[toán 10] chứng minh lượng giác

N

noinhobinhyen

1.

$sin^4x-sin^4(\dfrac{\pi}{2}-x)=sin^4x-cos^4x=sin^2x-cos^2x=2sin^2x-1$

2.

Ta có $cotA=\dfrac{b^2+c^2-a^2}{4S} ; cotB=\dfrac{a^2+c^2-b^2}{4S}$

$\Rightarrow \dfrac{tanA}{tanB}=\dfrac{cotB}{cotA}=\dfrac{a^2+c^2-b^2}{b^2+c^2-a^2}$


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$cotA = \dfrac{cosA}{sinA}=\dfrac{b^2+c^2-a^2}{2bc.sinA}=\dfrac{b^2+c^2-a^2}{4S}$
 
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