Toán Cho sin x + cos x = 1/5 . Tính | sin x − cos x | 11/07/2021 By Quinn Cho sin x + cos x = 1/5 . Tính | sin x − cos x |
$sinx+cosx=\frac{1}{5}$ $\Leftrightarrow sin^2x+cos^2x+2sinx.cosx=\frac{1}{25}$ $\Leftrightarrow 2sinx.cosx=\frac{1}{25}-1=\frac{-24}{25}$ $|sinx-cosx|=\sqrt{(sinx-cosx)^2}= \sqrt{sin^2x+cos^2x-2sinx.cosx}=\sqrt{1+\frac{24}{25}}=\frac{7}{5}$ Trả lời
$sinx+cosx=$$\frac{1}{5}$ ⇒ $(sinx+cosx)^2=$$(\frac{1}{5})^2$ ⇔ $sin^2x+cos^2+2sinxcosx=$$\frac{1}{25}$ ⇔ $1+2sinxcosx=$$\frac{1}{25}$ ( do $sin^2x+cos^2x=1$) ⇔ $2sinxcosx=$$\frac{1}{25}-1$ ⇔ $2sinxcosx=$$\frac{-24}{25}$ ⇔ $(sinx-cosx)^2=$$\frac{-24}{25}$ ⇔ $(sinx-cosx)^2=$$\frac{-24}{25}$ ⇔ $sin^2x+cos^2x-2sinxcosx=$$1-\frac{-24}{25}$ ⇔ $sin^2x+cos^2x-2sinxcosx=$$\frac{49}{25}$ ⇒ $| sin x − cos x |=$$\sqrt[]{(sinx-cosx)^2}=$ $\frac{7}{5}$ Trả lời
$sinx+cosx=\frac{1}{5}$
$\Leftrightarrow sin^2x+cos^2x+2sinx.cosx=\frac{1}{25}$
$\Leftrightarrow 2sinx.cosx=\frac{1}{25}-1=\frac{-24}{25}$
$|sinx-cosx|=\sqrt{(sinx-cosx)^2}= \sqrt{sin^2x+cos^2x-2sinx.cosx}=\sqrt{1+\frac{24}{25}}=\frac{7}{5}$
$sinx+cosx=$$\frac{1}{5}$
⇒ $(sinx+cosx)^2=$$(\frac{1}{5})^2$
⇔ $sin^2x+cos^2+2sinxcosx=$$\frac{1}{25}$
⇔ $1+2sinxcosx=$$\frac{1}{25}$ ( do $sin^2x+cos^2x=1$)
⇔ $2sinxcosx=$$\frac{1}{25}-1$
⇔ $2sinxcosx=$$\frac{-24}{25}$
⇔ $(sinx-cosx)^2=$$\frac{-24}{25}$
⇔ $(sinx-cosx)^2=$$\frac{-24}{25}$
⇔ $sin^2x+cos^2x-2sinxcosx=$$1-\frac{-24}{25}$
⇔ $sin^2x+cos^2x-2sinxcosx=$$\frac{49}{25}$
⇒ $| sin x − cos x |=$$\sqrt[]{(sinx-cosx)^2}=$ $\frac{7}{5}$