Question

Given f(x)= ln⁡(cos⁡(x)),calculate the value of ln⁡(cos⁡(^π⁄2)).

(a) -1.741

(b) 1.741

(c) 1.563

(d) -1.563

I got this question in exam.

Question is taken from Taylor Mclaurin Series topic in division Differential Calculus of Engineering Mathematics

Answer 1

Correct answer is (a) -1.741

To elaborate: Given f(x) = ln⁡(Cos(x)), f(0) = 0

Differentiating it f'(x) = – tan⁡(x), f'(0) = 0

Again \(f^{”}(x)=-sec^2⁡(x),f^{”}(0)=-1\)

And \(f^{”’}(x)=-2 sec⁡(x) \,sec⁡(x) \,tan⁡(x)=-2f^{”}(x)f'(x)\), hence f”(0)=0

f””(x)=-2(f”'(x) f'(x)+(f”(x))^2), hence f””(0)=-2

Now, by mclaurins’s series

f(x)=ln⁡(Cos(x))=\(0+0-x^2/2!+0-x^4/12+..\)

Therefore, f(x)=\(-x^2/2!-x^4/12+…\)

Hence,

ln⁡(cos⁡(π/2))=-1.741