Question

Find the value of ln(sin(31^o)) if ln(2) = 0.69315

(a) -0.653

(b) -0.663

(c) -0.764

(d) -0.662

The question was posed to me in an internship interview.

My doubt is from Taylor Mclaurin Series topic in section Differential Calculus of Engineering Mathematics

Answer 1

Right answer is (b) -0.663

For explanation: Let, f(x) = ln⁡(sin⁡(x+h))

Then, f(x) = ln⁡(sin⁡(x)), if h=0

f’ (x)=cot⁡(x), f” (x)=-cosec^2 (x), f”’ (x)=2cosec^2 (x)cot⁡(x)

Hence, by Taylor’s theorem,

f(x+h)=f(x)+hf'(x)+\(\frac{h^2}{2!}\) f” (x)+\(\frac{h^3}{3!}\) f”’ (x)+⋯..

Hence, ln⁡(sin⁡(x+h))=ln⁡(sin⁡(x))+h cot⁡(x)-\(\frac{h^2}{2!}\) cosec^2 (x)+\(\frac{h^3}{3!}\) (2cosec^2 (x)  cot⁡(x))+⋯.

Now let, x=30^o, h=1^o;

ln(sin(31^o)) = \(ln(sin(30))+\frac{π}{180} cot⁡(\frac{π}{6})-\frac{1}{2!} (\frac{π}{180})^2 cosec^2 (\frac{π}{6})+⋯\)

ln(sin(31^o)) = -0.6935+.030231-.000304+0

ln(sin(31^o)) = -0.663