Cho a,b nhọn và cosa = $\frac{1}{4}$ , cosb = $\frac{1}{5}$ Tính : sin(a+b) . sin(a-b) 06/08/2021 Bởi Katherine Cho a,b nhọn và cosa = $\frac{1}{4}$ , cosb = $\frac{1}{5}$ Tính : sin(a+b) . sin(a-b)
$A=\sin(a+b).\sin(a-b)$ $=\dfrac{-1}{2}(\cos 2a-\cos 2b)$ Mặt khác: $\cos^2a=\dfrac{1}{16}$ $\Rightarrow \sin^2a=1-\cos^2a=\dfrac{15}{16}$ $\Rightarrow \cos 2a=\cos^2a-\sin^2a=\dfrac{-7}{8}$ $\cos^2b=\dfrac{1}{25}$ $\Rightarrow \sin^2b=1-\cos^2b=\dfrac{24}{25}$ $\Rightarrow \cos 2b=\cos^2b-\sin^2b=\dfrac{-23}{25}$ $\to A=\dfrac{-1}{2}(\dfrac{-7}{8}+\dfrac{23}{25})=\dfrac{-9}{400}$ Bình luận
`sin a = \sqrt(1-(1/4)^2) = \sqrt15/4` `sin b = \sqrt(1-(1/5)^2) = (2\sqrt6)/5` `=> sin(a+b) = sinacosb + cosasinb` `= \sqrt15/4 . 1/5 + 1/4 . (2\sqrt6)/5 = (\sqrt15 + 2\sqrt6)/20` `sin (a-b) = sinacosb – cosasinb` `= sqrt15/4 . 1/5 – 1/4 . (2\sqrt6)/5 = (\sqrt15 – 2\sqrt6)/20` Bình luận
$A=\sin(a+b).\sin(a-b)$
$=\dfrac{-1}{2}(\cos 2a-\cos 2b)$
Mặt khác:
$\cos^2a=\dfrac{1}{16}$
$\Rightarrow \sin^2a=1-\cos^2a=\dfrac{15}{16}$
$\Rightarrow \cos 2a=\cos^2a-\sin^2a=\dfrac{-7}{8}$
$\cos^2b=\dfrac{1}{25}$
$\Rightarrow \sin^2b=1-\cos^2b=\dfrac{24}{25}$
$\Rightarrow \cos 2b=\cos^2b-\sin^2b=\dfrac{-23}{25}$
$\to A=\dfrac{-1}{2}(\dfrac{-7}{8}+\dfrac{23}{25})=\dfrac{-9}{400}$
`sin a = \sqrt(1-(1/4)^2) = \sqrt15/4`
`sin b = \sqrt(1-(1/5)^2) = (2\sqrt6)/5`
`=> sin(a+b) = sinacosb + cosasinb`
`= \sqrt15/4 . 1/5 + 1/4 . (2\sqrt6)/5 = (\sqrt15 + 2\sqrt6)/20`
`sin (a-b) = sinacosb – cosasinb`
`= sqrt15/4 . 1/5 – 1/4 . (2\sqrt6)/5 = (\sqrt15 – 2\sqrt6)/20`