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Question Number 34980 by NECx last updated on 14/May/18

If cos45° =(1/(√2)) . Find cos45.1°

$${If}\:{cos}\mathrm{45}°\:=\frac{\mathrm{1}}{\sqrt{\mathrm{2}}}\:.\:{Find}\:{cos}\mathrm{45}.\mathrm{1}° \\ $$

Answered by ajfour last updated on 14/May/18

cos ((π/4)+(π/(1800)))≈cos ((π/4))−(π/(1800))sin (π/4)      =(1/(√2))(1−(π/(1800))) =0.707(1−((0.0314)/(18)))     =0.707(1−0.00173)     ≈ 0.707−0.0012 ≈ 0.7058 .

$$\mathrm{cos}\:\left(\frac{\pi}{\mathrm{4}}+\frac{\pi}{\mathrm{1800}}\right)\approx\mathrm{cos}\:\left(\frac{\pi}{\mathrm{4}}\right)−\frac{\pi}{\mathrm{1800}}\mathrm{sin}\:\frac{\pi}{\mathrm{4}} \\ $$$$\:\:\:\:=\frac{\mathrm{1}}{\sqrt{\mathrm{2}}}\left(\mathrm{1}−\frac{\pi}{\mathrm{1800}}\right)\:=\mathrm{0}.\mathrm{707}\left(\mathrm{1}−\frac{\mathrm{0}.\mathrm{0314}}{\mathrm{18}}\right) \\ $$$$\:\:\:=\mathrm{0}.\mathrm{707}\left(\mathrm{1}−\mathrm{0}.\mathrm{00173}\right) \\ $$$$\:\:\:\approx\:\mathrm{0}.\mathrm{707}−\mathrm{0}.\mathrm{0012}\:\approx\:\mathrm{0}.\mathrm{7058}\:. \\ $$

Answered by tanmay.chaudhury50@gmail.com last updated on 14/May/18

y=cosx  y+△y=cos(x+△x) here x=45^(o )  x+△x=45.1^o   △y=cos(x+△x)−cosx     ((△y)/(△x))=(dy/dx)  △y=(dy/dx).△x  △y=−sinx×(Π/(1800))  △y=−((1/(√2)))×(Π/(1800))  so  cos(45.1^o )=(1/(√2))−(1/(√2))×(Π/(1800))  cos(45.1)^o =(1/(√2))(1−(Π/(1800)))  cos(45.1)^o =((√2)/2)×(1−0.0017453293)  cos(45.1)^o =0.7071067812×0.9982546707  cos(45.1)^o =0.705872647  180^o =Π radian  1^o =(Π/(180 ))   so  (0.1)^o =(Π/(1800))

$${y}={cosx} \\ $$$${y}+\bigtriangleup{y}={cos}\left({x}+\bigtriangleup{x}\right)\:{here}\:{x}=\mathrm{45}^{{o}\:} \:{x}+\bigtriangleup{x}=\mathrm{45}.\mathrm{1}^{{o}} \\ $$$$\bigtriangleup{y}={cos}\left({x}+\bigtriangleup{x}\right)−{cosx} \\ $$$$\:\:\:\frac{\bigtriangleup{y}}{\bigtriangleup{x}}=\frac{{dy}}{{dx}} \\ $$$$\bigtriangleup{y}=\frac{{dy}}{{dx}}.\bigtriangleup{x} \\ $$$$\bigtriangleup{y}=−{sinx}×\frac{\Pi}{\mathrm{1800}} \\ $$$$\bigtriangleup{y}=−\left(\frac{\mathrm{1}}{\sqrt{\mathrm{2}}}\right)×\frac{\Pi}{\mathrm{1800}} \\ $$$${so}\:\:{cos}\left(\mathrm{45}.\mathrm{1}^{{o}} \right)=\frac{\mathrm{1}}{\sqrt{\mathrm{2}}}−\frac{\mathrm{1}}{\sqrt{\mathrm{2}}}×\frac{\Pi}{\mathrm{1800}} \\ $$$${cos}\left(\mathrm{45}.\mathrm{1}\right)^{{o}} =\frac{\mathrm{1}}{\sqrt{\mathrm{2}}}\left(\mathrm{1}−\frac{\Pi}{\mathrm{1800}}\right) \\ $$$${cos}\left(\mathrm{45}.\mathrm{1}\right)^{{o}} =\frac{\sqrt{\mathrm{2}}}{\mathrm{2}}×\left(\mathrm{1}−\mathrm{0}.\mathrm{0017453293}\right) \\ $$$${cos}\left(\mathrm{45}.\mathrm{1}\right)^{{o}} =\mathrm{0}.\mathrm{7071067812}×\mathrm{0}.\mathrm{9982546707} \\ $$$${cos}\left(\mathrm{45}.\mathrm{1}\right)^{{o}} =\mathrm{0}.\mathrm{705872647} \\ $$$$\mathrm{180}^{{o}} =\Pi\:{radian} \\ $$$$\mathrm{1}^{{o}} =\frac{\Pi}{\mathrm{180}\:}\:\:\:{so}\:\:\left(\mathrm{0}.\mathrm{1}\right)^{{o}} =\frac{\Pi}{\mathrm{1800}} \\ $$$$ \\ $$

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