Question

the approximate value of sin(31°) given that 1°=0.0175, cos 30°=0.8660 is​

Answer 1

$\large\underline{\sf{Solution-}}$

Let assume that

$\rm :\longmapsto\:f(x) = sinx \: \: \: where \: x \: = \: 30 \degree$

So,

$\rm :\longmapsto\:f(x + \triangle x) = sin(x + \triangle x) \: \: \: where \: \triangle x \: = \: 1 \degree$

Using definition of differentiation, we have

$\rm :\longmapsto\:f(x + \triangle x) = f(x) + f'(x)\triangle x$

On substituting the values of

$\rm :\longmapsto\:f(x) = sinx$

$\rm :\longmapsto\:f'(x) = cosx$

So, we have .

$\rm :\longmapsto\:sin(x + \triangle x) = sinx + cosx \times \triangle x$

So, we have now on substituting the values

$\rm :\longmapsto\:sin(30\degree + 1\degree) = sin30\degree + cos30\degree \times 1\degree$

$\rm :\longmapsto\:sin31\degree = 0.5 +0.8660 \times 0.0175$

$\rm :\longmapsto\:sin31\degree = 0.5 +0.0015$

$\rm :\longmapsto\:sin31\degree = 0.5015$

Hence,

$\purple{\rm\implies \:\boxed{\tt{ \: \: sin31\degree = 0.5015 \: \: }}}$

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LEARN MORE

$\begin{gathered}\boxed{\begin{array}{c|c} \bf f(x) & \bf \dfrac{d}{dx}f(x) \\ \\ \frac{\qquad \qquad}{} & \frac{\qquad \qquad}{} \\ \sf k & \sf 0 \\ \\ \sf sinx & \sf cosx \\ \\ \sf cosx & \sf - \: sinx \\ \\ \sf tanx & \sf {sec}^{2}x \\ \\ \sf cotx & \sf - {cosec}^{2}x \\ \\ \sf secx & \sf secx \: tanx\\ \\ \sf cosecx & \sf - \: cosecx \: cotx\\ \\ \sf \sqrt{x} & \sf \dfrac{1}{2 \sqrt{x} } \\ \\ \sf logx & \sf \dfrac{1}{x}\\ \\ \sf {e}^{x} & \sf {e}^{x} \end{array}} \\ \end{gathered}$

Answer 2

Answer:

$\large\underline{\sf{Solution-}}$

Let assume that

$\rm :\longmapsto\:f(x) = sinx \: \: \: where \: x \: = \: 30 \degree$

So,

$\rm :\longmapsto\:f(x + \triangle x) = sin(x + \triangle x) \: \: \: where \: \triangle x \: = \: 1 \degree$

Using definition of differentiation, we have

$\rm :\longmapsto\:f(x + \triangle x) = f(x) + f'(x)\triangle x$

On substituting the values of

$\rm :\longmapsto\:f(x) = sinx$

$\rm :\longmapsto\:f'(x) = cosx$

So, we have .

$\rm :\longmapsto\:sin(x + \triangle x) = sinx + cosx \times \triangle x$

So, we have now on substituting the values

$\rm :\longmapsto\:sin(30\degree + 1\degree) = sin30\degree + cos30\degree \times 1\degree$

$\rm :\longmapsto\:sin31\degree = 0.5 +0.8660 \times 0.0175$

$\rm :\longmapsto\:sin31\degree = 0.5 +0.0015$

$\rm :\longmapsto\:sin31\degree = 0.5015$

Hence,

$\pink{\sf\implies \:\boxed{\tt{ \: \: sin31\degree = 0.5015 \: \: }}}$

▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬

LEARN MORE

$\begin{gathered}\boxed{\begin{array}{c|c} \bf f(x) & \bf \dfrac{d}{dx}f(x) \\ \\ \frac{\qquad \qquad}{} & \frac{\qquad \qquad}{} \\ \sf k & \sf 0 \\ \\ \sf sinx & \sf cosx \\ \\ \sf cosx & \sf - \: sinx \\ \\ \sf tanx & \sf {sec}^{2}x \\ \\ \sf cotx & \sf - {cosec}^{2}x \\ \\ \sf secx & \sf secx \: tanx\\ \\ \sf cosecx & \sf - \: cosecx \: cotx\\ \\ \sf \sqrt{x} & \sf \dfrac{1}{2 \sqrt{x} } \\ \\ \sf logx & \sf \dfrac{1}{x}\\ \\ \sf {e}^{x} & \sf {e}^{x} \end{array}} \\ \end{gathered}$