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Question Number 147031 by mnjuly1970 last updated on 17/Jul/21

Answered by mindispower last updated on 17/Jul/21

$$\left(1\right)+\left(2\right)\Rightarrow cos\left(x\right)+sin\left(x\right)-\sqrt{2}=\frac{cos\left(2y\right)}{\sqrt{2}}$$$$\Rightarrow \sqrt{2}cos(\frac{\pi }{4}-x)-\sqrt{2}=\frac{cos\left(2y\right)}{\sqrt{2}}$$$$\Rightarrow cos\left(2y\right)=2cos(\frac{\pi }{4}-x)-2$$$$$$

Commented by mnjuly1970 last updated on 17/Jul/21

$$thanksalot$$

Answered by gsk2684 last updated on 17/Jul/21

$$\cos x-\sin x=\frac{1}{\sqrt{2}}$$$$\cos (x+\frac{\pi }{4})=\frac{1}{2}$$$$x+\frac{\pi }{4}=2n\pi \pm \frac{\pi }{3},n\in Z$$$$x=2n\pi +\frac{\pi }{12},2n\pi -\frac{7\pi }{12}wheren\in Z$$$$and$$$$\cos x+\sin x=\frac{1}{\sqrt{2}}(2+\cos 2y)$$$$case\mathrm{i}:$$$$\frac{\sqrt{3}+1}{2\sqrt{2}}+\frac{\sqrt{3}-1}{2\sqrt{2}}=\frac{1}{\sqrt{2}}(2+\cos 2\mathrm{y})$$$$\frac{\sqrt{3}}{\sqrt{2}}=\frac{1}{\sqrt{2}}(2+\cos 2\mathrm{y})$$$$\cos 2y=\sqrt{3}-2$$$$caseii:$$$$-\frac{\sqrt{3}-1}{2\sqrt{2}}-\frac{\sqrt{3}+1}{2\sqrt{2}}=\frac{1}{\sqrt{2}}(2+\cos 2y)$$$$\frac{-\sqrt{3}}{\sqrt{2}}=\frac{1}{\sqrt{2}}(2+\cos 2y)$$$$\cos 2y=-\sqrt{3}-2$$$$absurdity$$

Commented by mnjuly1970 last updated on 17/Jul/21

$$grateful...$$

Answered by mr W last updated on 17/Jul/21

$$\left(i\right)-\left(ii\right):$$$$\cos x-\sin x=\frac{1}{\sqrt{2}}$$$$\left(i\right)+\left(ii\right):$$$$\cos x+\sin x=\frac{2+\cos 2y}{\sqrt{2}}$$$$\Rightarrow \cos x=\frac{3+\cos 2y}{2\sqrt{2}}$$$$\Rightarrow \sin x=\frac{1+\cos 2y}{2\sqrt{2}}$$$${\left(\frac{3+\cos 2y}{2\sqrt{2}}\right)}^2+{\left(\frac{1+\cos 2y}{2\sqrt{2}}\right)}^2=1$$$${\cos }^{2}2y+4\cos 2y+1=0$$$$\cos 2y=\frac{-4\pm 2\sqrt{3}}{2}=-2+\sqrt{3},-2-\sqrt{3}\left(rejected\right)$$

Commented by mnjuly1970 last updated on 17/Jul/21

$$thxmr\mathrm{W}$$

Answered by EDWIN88 last updated on 17/Jul/21

$$\sin {}^2\mathrm{x}+\cos {}^2\mathrm{x}=1$$$${\left(\frac{\cos {}^2\mathrm{y}}{\sqrt{2}}\right)}^2+{\left(\frac{1+\cos {}^2\mathrm{y}}{\sqrt{2}}\right)}^2=1$$$${\left(\frac{1+\cos 2\mathrm{y}}{2}\right)}^2+{(1+\frac{1+\cos 2\mathrm{y}}{2})}^2=2$$$${(1+\cos 2\mathrm{y})}^2+{(3+\cos 2\mathrm{y})}^2=8$$$$2\cos {}^{2}2\mathrm{y}+8\cos 2\mathrm{y}+2=0$$$$\Rightarrow \cos {}^{2}2\mathrm{y}+4\cos 2\mathrm{y}+1=0$$$$\Rightarrow \cos 2\mathrm{y}=\frac{-4+\sqrt{12}}{2}=-2+\sqrt{3}$$$$$$

Commented by mnjuly1970 last updated on 17/Jul/21

$$gratefulmredwin$$

Answered by behi834171 last updated on 17/Jul/21

$$\sqrt{2}(cosx-sinx)=1\Rightarrow cos(x+\frac{\pi }{4})=\frac{1}{2}\Rightarrow$$$$x+\frac{\pi }{4}=\frac{\pi }{3}or\frac{2\pi }{3}\Rightarrow x=\frac{\pi }{12}or\frac{5\pi }{12}$$$$\Rightarrow {cos}^{2}y=\sqrt{2}sinx=\sqrt{2}\times \frac{\sqrt{6}\pm \sqrt{2}}{4}=\frac{\sqrt{3}\pm 1}{2}$$$$\mathit{c}\mathit{o}\mathit{s}2\mathit{y}=2{\mathit{c}\mathit{o}\mathit{s}}^2\mathit{y}-1=(\sqrt{3}\pm 1)-1=\left(\sqrt{3}\right)or(\sqrt{3}-2)$$$$\Rightarrow \mathit{c}\mathit{o}\mathit{s}2\mathit{y}=(\sqrt{3}-2).◼$$

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