Using differentials, find the approximate value of ,[tex]\pmb{\tt\cos \;\; 29°}[/tex]upto three places of decimal.
[tex] \rule{300pt}{0.1pt}[/tex]
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[tex]\large\underline{\sf{Solution-}}[/tex]
Let assume that
[tex]\sf \: f(x) = cosx, \: \: \: where \: x = 30 \degree \: \\ \\ [/tex]
Now,
[tex]\sf \: f(x + \triangle x\: ) = cos(x + \triangle x), \: \: \: where \: x \: = \: - \: 1 \degree \: \\ \\ [/tex]
Using Definition of differentials, we have
[tex]\sf \: f(x + \triangle x\:) = f(x) + f'(x) \: \triangle x\: \\ \\ [/tex]
[tex]\sf \: cos(x + \triangle x\:) = cosx + ( - \: sinx) \: \triangle x\: \\ \\ [/tex]
[tex]\sf \: cos(x + \triangle x\:) = cosx \: - \: sinx\: \triangle x\: \\ \\ [/tex]
On substituting the values, we get
[tex]\sf \: cos(30\degree - 1\degree ) = cos30\degree \: - \: sin30\degree \: \times (1\degree )\: \\ \\ [/tex]
[tex]\sf \: cos29\degree = \dfrac{ \sqrt{3} }{2} - \dfrac{1}{2} \times 0.01745 \\ \\ [/tex]
[tex]\sf \: cos29\degree = 0.866 - 0.008725 \\ \\ [/tex]
[tex]\bf\implies \:cos29\degree \: = \: 0.857275 \\ \\ [/tex]
[tex]\rule{190pt}{2pt}[/tex]
[tex] {{ \mathfrak{Additional\:Information}}}[/tex]
[tex]\begin{gathered}\boxed{\begin{array}{c|c} \bf f(x) & \bf \dfrac{d}{dx}f(x) \\ \\ \frac{\qquad \qquad}{} & \frac{\qquad \qquad}{} \\ \sf k & \sf 0 \\ \\ \sf sinx & \sf cosx \\ \\ \sf cosx & \sf - \: sinx \\ \\ \sf tanx & \sf {sec}^{2}x \\ \\ \sf cotx & \sf - {cosec}^{2}x \\ \\ \sf secx & \sf secx \: tanx\\ \\ \sf cosecx & \sf - \: cosecx \: cotx\\ \\ \sf \sqrt{x} & \sf \dfrac{1}{2 \sqrt{x} } \\ \\ \sf logx & \sf \dfrac{1}{x}\\ \\ \sf {x}^{n} & \sf {nx}^{n - 1}\\ \\ \sf {e}^{x} & \sf {e}^{x} \end{array}} \\ \end{gathered}[/tex]
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