Question

Approximate value of cos-1(-0.49) is ________
A. $\dfrac{{2\pi }}{3}$ +$\dfrac{1}{{50(\sqrt 3 )}}$
B. $\dfrac{{2\pi }}{3}$ - $\dfrac{1}{{50(\sqrt 3 )}}$
C. $\dfrac{\pi }{3}$ - $\dfrac{1}{{50(\sqrt 3 )}}$
D. $\dfrac{\pi }{3}$ +$\dfrac{1}{{50(\sqrt 3 )}}$

Answer 1

Hint: This is a simple problem of error and approximation. We have to know the concept of differentiation and limits.
It may be solved by knowing the concept of applications of derivatives.We can use differentials to calculate small changes in the dependent variable of a function corresponding to small changes in the independent variable. The theory behind it is quite simple: From the chapter on applications of derivatives
As we know that, $\dfrac{{dy}}{{dx}}$ = $\mathop {\lim }\limits_{\delta x \to 0} \dfrac{{f(x + h) - f(x)}}{h}$ ……….eq(1)
So, f(x+h) = f(x) + h$\dfrac{{dy}}{{dx}}$ ………..eq(2)

Complete step-by-step answer:
Given, f(x)= cos-1(x)
We have to find the value of cos-1(-0.49).
It can be written as ,-0.49= -0.5 + 0.01
So, cos-1(-0.49) = cos-1(-0.5 + 0.01)= f(x+h)
Then, x= -0.5 and h= 0.01
Now, we use the formula ,
$\dfrac{{dy}}{{dx}}$ = $\mathop {\lim }\limits_{\delta x \to 0} \dfrac{{f(x + h) - f(x)}}{h}$ and f(x+h) = f(x) + h$\dfrac{{dy}}{{dx}}$
$\dfrac{{dy}}{{dx}}$ = $\dfrac{{d(co{s^{ - 1}}x)}}{{dx}}$ = \[\dfrac{{ - 1}}{{\sqrt {1 - {x^2}} }}\]
STEP 1-
And putting x= -0.5 we get, $\dfrac{{dy}}{{dx}}$= \[\dfrac{{ - 1}}{{\sqrt {1 - {{( - 0.5)}^2}} }}\] = \[\dfrac{{ - 1}}{{\sqrt {0.75} }}\]= \[\dfrac{{ - 2}}{{\sqrt 3 }}\]
STEP 2 -
f(x)= cos-1(x) , so at x= -0.5 ,f(x)= cos-1(-0.5)= $\dfrac{{2\pi }}{3}$
STEP 3-
Finally using the formulae i.e. f(x+h) = f(x) + h$\dfrac{{dy}}{{dx}}$
cos-1(-0.49) = $\dfrac{{2\pi }}{3}$+ (0.01×\[\dfrac{{ - 2}}{{\sqrt 3 }}\])= $\dfrac{{2\pi }}{3}$- $\dfrac{1}{{50\sqrt 3 }}$
Hence, we finally got the answer that is $\dfrac{{2\pi }}{3}$- $\dfrac{1}{{50\sqrt 3 }}$

OPTION (B) IS CORRECT

Note:
One should be very aware when converting or differentiating inverse cosine function i.e. f(x)= cos-1(x)
As it is $\dfrac{{dy}}{{dx}}$ = $\dfrac{{d(co{s^{ - 1}}x)}}{{dx}}$ = \[\dfrac{{ - 1}}{{\sqrt {1 - {x^2}} }}\] (negative sign should be taken)
As it is similar to that of inverse sine function . so. Students may find it to be the same as it.