cho góc nhọn a sao cho sin a . cos a =1/3 tính giá trị của a =sin^6 a +cos^6 a 20/09/2021 Bởi Mary cho góc nhọn a sao cho sin a . cos a =1/3 tính giá trị của a =sin^6 a +cos^6 a
Ta có $\sin^6a + \cos^6a = (\sin^2a + \cos^2a)(\sin^4a + \cos^4a – \sin^2a \cos^2a)$ $= 1.(\sin^4a + \cos^4a – \dfrac{1}{9})$ $= (\sin^2a)^2 + (\cos^2a)^2 + 2\sin^2a \cos^2a – 2\sin^2a \cos^2a – \dfrac{1}{9}$ $= (\sin^2a + \cos^2a)^2 – 2\sin^2a \cos^2a – \dfrac{1}{9}$ $= 1 – 2.\dfrac{1}{9} – \dfrac{1}{9} = \dfrac{2}{3}$. Bình luận
$sin^6a+cos^6a$ $= (sin^2a)^3 + (cos^2a)^3$ $= (sin^2a+cos^2a)(sin^4a-sin^2a.cos^2a+cos^4a)$ $= sin^4a-sin^2a.cos^2a+cos^4x$ $= (sin^2a+cos^2a)^2-2sin^2a.cos^2a-sin^2a.cos^2a$ $= 1-3sin^2a.cos^2a$ $=1-3.(\frac{1}{3})^2$ $=\frac{2}{3}$ Bình luận
Ta có
$\sin^6a + \cos^6a = (\sin^2a + \cos^2a)(\sin^4a + \cos^4a – \sin^2a \cos^2a)$
$= 1.(\sin^4a + \cos^4a – \dfrac{1}{9})$
$= (\sin^2a)^2 + (\cos^2a)^2 + 2\sin^2a \cos^2a – 2\sin^2a \cos^2a – \dfrac{1}{9}$
$= (\sin^2a + \cos^2a)^2 – 2\sin^2a \cos^2a – \dfrac{1}{9}$
$= 1 – 2.\dfrac{1}{9} – \dfrac{1}{9} = \dfrac{2}{3}$.
$sin^6a+cos^6a$
$= (sin^2a)^3 + (cos^2a)^3$
$= (sin^2a+cos^2a)(sin^4a-sin^2a.cos^2a+cos^4a)$
$= sin^4a-sin^2a.cos^2a+cos^4x$
$= (sin^2a+cos^2a)^2-2sin^2a.cos^2a-sin^2a.cos^2a$
$= 1-3sin^2a.cos^2a$
$=1-3.(\frac{1}{3})^2$
$=\frac{2}{3}$