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Question Number 89193 by jagoll last updated on 16/Apr/20

$$\cos x-\sin x=\frac{1}{2}$$$$\cos x\sin x=\frac{3}{8},\pi <x<2\pi$$$$\cos x+\sin x=?$$

Commented by Tony Lin last updated on 16/Apr/20

$$\because \pi <x<2\pi$$$$\therefore sinx<0$$$$\because cosxsinx=\frac{3}{8}$$$$\therefore cosx<0$$$${(cosx+sinx)}^2={(cosx-sinx)}^2+4cosxsinx$$$$=\frac{1}{4}+\frac{3}{2}=\frac{7}{4}$$$$\Rightarrow cosx+sinx=-\frac{\sqrt{7}}{2}$$$$\{\begin{array}{l}cosx+sinx=-\frac{\sqrt{7}}{2}\\ cosx-sinx=\frac{1}{2}\end{array}$$$$\Rightarrow \{\begin{array}{l}sinx=\frac{-\sqrt{7}-1}{4}\\ cosx=\frac{-\sqrt{7}+1}{4}\end{array}$$

Commented by john santu last updated on 16/Apr/20

$$let\cos x+\sin x=p,p<0$$$$\frac{1}{2}p=\cos 2x\left(i\right)$$$$\left(ii\right)2\sin x\cos x=\frac{3}{4}\Rightarrow \sin 2x=\frac{3}{4}$$$$\cos 2x=-\frac{\sqrt{7}}{4}$$$$\therefore \frac{1}{2}p=-\frac{\sqrt{7}}{4}\Rightarrow p=-\frac{\sqrt{7}}{2}$$$$\cos x+\sin x=-\frac{\sqrt{7}}{2}$$

Commented by jagoll last updated on 16/Apr/20

$$thankyou$$

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