Cho sin a+cosa=1,2, tính sina.cosa và tg^3a+cot^3a November 15, 2020 by Ladonna Cho sin a+cosa=1,2, tính sina.cosa và tg^3a+cot^3a
$\sin a+\cos a=1,2$ $\Leftrightarrow \sin^2a+\cos^2a+2\sin a.\cos a=1,2^2=1,44$ $\Leftrightarrow \sin a.\cos a=\dfrac{1-1,44}{2}=-0,22$ $\tan^3a+\cot^3a$ $=\dfrac{\sin^3a}{\cos^3a}+\dfrac{\cos^3a}{\sin^3a}$ $=\dfrac{\sin^6a+\cos^6a}{\sin^3a.\cos^3a}$ $=\dfrac{(\sin^2a+\cos^2a)(\sin^4a+\cos^4a-\sin^2a\cos^2a)}{(\sin a.\cos a)^3}$ $=\dfrac{(\sin^2a+\cos^2a)^2-2\sin^2a\cos^2a}{(\sin a.\cos a)^3}$ $=\dfrac{1-3(\sin a\cos a)^2}{(\sin a\cos a)^3}$ $=\dfrac{1-3.0,22^2}{(-0,22)^3}\approx 80,28$ Reply
$\sin a+\cos a=1,2$
$\Leftrightarrow \sin^2a+\cos^2a+2\sin a.\cos a=1,2^2=1,44$
$\Leftrightarrow \sin a.\cos a=\dfrac{1-1,44}{2}=-0,22$
$\tan^3a+\cot^3a$
$=\dfrac{\sin^3a}{\cos^3a}+\dfrac{\cos^3a}{\sin^3a}$
$=\dfrac{\sin^6a+\cos^6a}{\sin^3a.\cos^3a}$
$=\dfrac{(\sin^2a+\cos^2a)(\sin^4a+\cos^4a-\sin^2a\cos^2a)}{(\sin a.\cos a)^3}$
$=\dfrac{(\sin^2a+\cos^2a)^2-2\sin^2a\cos^2a}{(\sin a.\cos a)^3}$
$=\dfrac{1-3(\sin a\cos a)^2}{(\sin a\cos a)^3}$
$=\dfrac{1-3.0,22^2}{(-0,22)^3}\approx 80,28$