Question and Answers Forum

All Questions      Topic List

Trigonometry Questions

Previous in All Question      Next in All Question      

Previous in Trigonometry      Next in Trigonometry      

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

Answered by mindispower last updated on 17/Jul/21

(1)+(2)⇒cos(x)+sin(x)−(√2)=((cos(2y))/( (√2))) ⇒(√2)cos((π/4)−x)−(√2)=((cos(2y))/( (√2))) ⇒cos(2y)=2cos((π/4)−x)−2

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

Commented by mnjuly1970 last updated on 17/Jul/21

thanks alot

$${thanks}\:{alot} \\ $$

Answered by gsk2684 last updated on 17/Jul/21

cos x−sin x=(1/( (√2))) cos (x+(π/4))=(1/2) x+(π/4)=2nπ±(π/3),n∈Z x=2nπ+(π/(12)),2nπ−((7π)/(12)) where n∈Z and cos x+sin x=(1/( (√2)))(2+cos 2y) case i: (((√3)+1)/(2(√2)))+(((√3)−1)/(2(√2)))=(1/( (√2)))(2+cos 2y) ((√3)/( (√2)))=(1/( (√2)))(2+cos 2y) cos 2y=(√3)−2 case ii: −(((√3)−1)/(2(√2)))−(((√3)+1)/(2(√2)))=(1/( (√2)))(2+cos 2y) ((−(√3))/( (√2)))=(1/( (√2)))(2+cos 2y) cos 2y=−(√3)−2 absurdity

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

Commented by mnjuly1970 last updated on 17/Jul/21

grateful ...

$${grateful}\:... \\ $$

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

(i)−(ii): cos x−sin x=(1/( (√2))) (i)+(ii): cos x+sin x=((2+cos 2y)/( (√2))) ⇒cos x=((3+cos 2y)/(2(√2))) ⇒sin x=((1+cos 2y)/(2(√2))) (((3+cos 2y)/(2(√2))))^2 +(((1+cos 2y)/(2(√2))))^2 =1 cos^2 2y+4cos 2y+1=0 cos 2y=((−4±2(√3))/2)=−2+(√3),−2−(√3) (rejected)

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

Commented by mnjuly1970 last updated on 17/Jul/21

thx mr W

$$\:{thx}\:{mr}\:\mathrm{W} \\ $$

Answered by EDWIN88 last updated on 17/Jul/21

sin^2 x+cos^2 x = 1 (((cos^2 y)/( (√2))))^2 +(((1+cos^2 y)/( (√2))))^2 = 1 (((1+cos 2y)/2))^2 +(1+((1+cos 2y)/2))^2 = 2 (1+cos 2y)^2 +(3+cos 2y)^2 = 8 2cos^2 2y + 8cos 2y+2 = 0 ⇒cos^2 2y+4cos 2y+1=0 ⇒cos 2y = ((−4+(√(12)))/2)=−2+(√3)

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

Commented by mnjuly1970 last updated on 17/Jul/21

grateful mr edwin

$${grateful}\:{mr}\:{edwin} \\ $$

Answered by behi834171 last updated on 17/Jul/21

(√2)(cosx−sinx)=1⇒cos(x+(π/4))=(1/2)⇒ x+(π/4)=(π/3) or ((2π)/3)⇒x=(π/(12)) or ((5π)/(12)) ⇒cos^2 y=(√2)sinx=(√2)×(((√6)±(√2))/4)=(((√3)±1)/2) cos2y=2cos^2 y−1=((√3)±1)−1=((√3)) or((√3)−2) ⇒cos2y=((√3)−2) .■

$$\sqrt{\mathrm{2}}\left({cosx}−{sinx}\right)=\mathrm{1}\Rightarrow{cos}\left({x}+\frac{\pi}{\mathrm{4}}\right)=\frac{\mathrm{1}}{\mathrm{2}}\Rightarrow \\ $$$${x}+\frac{\pi}{\mathrm{4}}=\frac{\pi}{\mathrm{3}}\:{or}\:\frac{\mathrm{2}\pi}{\mathrm{3}}\Rightarrow{x}=\frac{\pi}{\mathrm{12}}\:{or}\:\frac{\mathrm{5}\pi}{\mathrm{12}} \\ $$$$\Rightarrow{cos}^{\mathrm{2}} {y}=\sqrt{\mathrm{2}}{sinx}=\sqrt{\mathrm{2}}×\frac{\sqrt{\mathrm{6}}\pm\sqrt{\mathrm{2}}}{\mathrm{4}}=\frac{\sqrt{\mathrm{3}}\pm\mathrm{1}}{\mathrm{2}} \\ $$$$\boldsymbol{{cos}}\mathrm{2}\boldsymbol{{y}}=\mathrm{2}\boldsymbol{{cos}}^{\mathrm{2}} \boldsymbol{{y}}−\mathrm{1}=\left(\sqrt{\mathrm{3}}\pm\mathrm{1}\right)−\mathrm{1}=\left(\sqrt{\mathrm{3}}\right)\:{or}\left(\sqrt{\mathrm{3}}−\mathrm{2}\right) \\ $$$$\Rightarrow\boldsymbol{{cos}}\mathrm{2}\boldsymbol{{y}}=\left(\sqrt{\mathrm{3}}−\mathrm{2}\right)\:\:\:.\blacksquare \\ $$

Terms of Service

Privacy Policy

Contact: info@tinkutara.com