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

If sin (45°−x) = −(1/3) , where 45°<x<90° .What is the value of (6sin x−(√2))^2

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

Answered by bemath last updated on 04/Oct/20

⇒sin (x−45) = (1/3) ⇒((√2)/2) sin x−((√2)/2) cos x = (1/3) ⇒(√2) sin x − (√2) cos x = (2/3) ⇒ sin x−cos x = ((√2)/3) ⇒ sin x−((√2)/3) = (√(1−sin^2 x)) ⇒sin^2 x−((2(√2))/3) sin x +(2/9) = 1−sin^2 x ⇒2 sin^2 x−((2(√2))/3) sin x−(7/9)=0 ⇒sin x = ((((2(√2))/3) + (√((8/9)+((56)/9))))/4) ⇒ sin x = ((2(√2) +8)/(12)) = ((4+(√2))/6) ⇒6 sin x = 4+(√2) ⇒(6sin x−(√2))^2 = 16

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

Commented bybobhans last updated on 04/Oct/20

cooll

$$\mathrm{cooll} \\ $$

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

(1/( (√2)))cosx−(1/( (√2)))sinx=−(1/3) sinx−cosx=((√2)/3) sin^2 x−2((√2)/3)sinx+(2/9)=cos^2 x 2sin^2 x−((2(√2))/3)sinx−(7/9)=0 sinx=((((2(√2))/3)±(√((8/9)+((56)/9))))/4)=((((√2)/3)±(4/3))/2)=((4±(√2))/6) (6sinx−(√2))^2 =4^2 =16 (As (π/4)<x<(π/2))

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

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