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Question Number 116441 by bobhans last updated on 04/Oct/20

$$\mathrm{If}\sin (45°-\mathrm{x})=-\frac{1}{3},\mathrm{where}45°<\mathrm{x}<90°$$ $$.\mathrm{What}\mathrm{is}\mathrm{the}\mathrm{value}\mathrm{of}{(6\sin \mathrm{x}-\sqrt{2})}^2$$

Answered by bemath last updated on 04/Oct/20

$$\Rightarrow \sin (\mathrm{x}-45)=\frac{1}{3}$$ $$\Rightarrow \frac{\sqrt{2}}{2}\sin \mathrm{x}-\frac{\sqrt{2}}{2}\cos \mathrm{x}=\frac{1}{3}$$ $$\Rightarrow \sqrt{2}\sin \mathrm{x}-\sqrt{2}\cos \mathrm{x}=\frac{2}{3}$$ $$\Rightarrow \sin \mathrm{x}-\cos \mathrm{x}=\frac{\sqrt{2}}{3}$$ $$\Rightarrow \sin \mathrm{x}-\frac{\sqrt{2}}{3}=\sqrt{1-\sin {}^2\mathrm{x}}$$ $$\Rightarrow \sin {}^2\mathrm{x}-\frac{2\sqrt{2}}{3}\sin \mathrm{x}+\frac{2}{9}=1-\sin {}^2\mathrm{x}$$ $$\Rightarrow 2\sin {}^2\mathrm{x}-\frac{2\sqrt{2}}{3}\sin \mathrm{x}-\frac{7}{9}=0$$ $$\Rightarrow \sin \mathrm{x}=\frac{\frac{2\sqrt{2}}{3}+\sqrt{\frac{8}{9}+\frac{56}{9}}}{4}$$ $$\Rightarrow \sin \mathrm{x}=\frac{2\sqrt{2}+8}{12}=\frac{4+\sqrt{2}}{6}$$ $$\Rightarrow 6\sin \mathrm{x}=4+\sqrt{2}$$ $$\Rightarrow {(6\sin \mathrm{x}-\sqrt{2})}^2=16$$

Commented bybobhans last updated on 04/Oct/20

$$\mathrm{cooll}$$

Answered by Dwaipayan Shikari last updated on 04/Oct/20

$$\frac{1}{\sqrt{2}}\mathrm{cosx}-\frac{1}{\sqrt{2}}\mathrm{sinx}=-\frac{1}{3}$$ $$\mathrm{sinx}-\mathrm{cosx}=\frac{\sqrt{2}}{3}$$ $${\sin }^2\mathrm{x}-2\frac{\sqrt{2}}{3}\mathrm{sinx}+\frac{2}{9}={\cos }^2\mathrm{x}$$ $${2\sin }^2\mathrm{x}-\frac{2\sqrt{2}}{3}\mathrm{sinx}-\frac{7}{9}=0$$ $$\mathrm{sinx}=\frac{\frac{2\sqrt{2}}{3}\pm \sqrt{\frac{8}{9}+\frac{56}{9}}}{4}=\frac{\frac{\sqrt{2}}{3}\pm \frac{4}{3}}{2}=\frac{4\pm \sqrt{2}}{6}$$ $${(6\mathrm{sinx}-\sqrt{2})}^2=4^2=16(\mathrm{As}\frac{\pi }{4}<\mathrm{x}<\frac{\pi }{2})$$ $$$$

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