Question

How do you use a linear approximation to estimate $\sin {{28}^{\circ }}$?

Answer 1

In this problem, we have to use linear approximation to estimate $\sin {{28}^{\circ }}$. In this problem, we can use Taylor series expansion to find the approximation of the given problem. We already know the value of sin 30 degree is 0.5, we can apply this in Taylor series to get the approximate value of the given problem. We should know that to solve these types of problems, we have to know differentiation rules and formulas and we have to know trigonometric rules, degree values and formulas.

Complete step by step answer:
We know that the given degree to be estimated approximately is,
$\sin {{28}^{\circ }}$
We know that the Taylor series expansion is,
$f\left( x \right)=f\left( {{x}_{0}} \right)+f'\left( {{x}_{0}} \right)\left( x-{{x}_{0}} \right)+{{O}_{2}}\left( x-{{x}_{0}} \right)$
We also know that $\sin {{30}^{\circ }}=0.5$.
We know that ${{O}_{2}}\left( x-{{x}_{0}} \right)\approx 0$.
We can apply the Taylor series, to get the approximate value.
We know that if we differentiate the sine value, we get
$\sin xdx=-\cos x$

& \Rightarrow \sin {{28}^{\circ }}\approx \sin {{30}^{\circ }}+\cos {{30}^{\circ }}\left( \dfrac{28}{180}\pi -\dfrac{\pi }{6} \right) \\\ & \Rightarrow \sin {{28}^{\circ }}\approx \dfrac{1}{2}-\dfrac{\pi }{60\sqrt{3}} \\\ \end{aligned}$$ $$\Rightarrow \sin {{28}^{\circ }}\approx 0.46977$$ **Note:** Students make mistakes while using Taylor series expansion, in which we have to use values, differentiation, and finally add the terms. We should know that to solve these types of problems, we have to know differentiation rules and formulas and we have to know trigonometric rules, degree values and formulas. We can also use scientific calculators to check the values, we have worked out to be correct. It is better to use calculators to find the exact value for these types of problems.