Question

If $a,b$ are positive quantities and if
${a_1} = \dfrac{{a + b}}{2}$ , ${b_1} = \sqrt {{a_1}{b_1}}$
${a_2} = \dfrac{{a{}_1 + {b_1}}}{2}$ , ${b_2} = \sqrt {{a_2}{b_2}}$
A) ${b_\infty } = \dfrac{{\sqrt {{a^2} + {b^2}} }}{{{{\cos }^{ - 1}}(a/b)}}$
B) ${b_\infty } = \dfrac{{\sqrt {{a^2} - {b^2}} }}{{{{\cos }^{ - 1}}(a/b)}}$
C) ${b_\infty } = \dfrac{{\sqrt {({a^2} - {b^2})} }}{{{{\cos }^{ - 1}}(a/b)}}$
D) None of these

Answer 1

To solve this question firstly we consider a particular value of $'a'$ ‘a’ in the form of $b$and$\cos \theta$. With the help of this value we can find the value of ${a_1}$ as we have given${a_1} = \dfrac{{a + b}}{2}$. Once value of ${b_1}$. We can calculate the value of ${b_1}$in the form of $\cos \theta$ function. With the help of the value of ${a_1}$ and ${b_1}$ we can calculate the value of ${b_2}$ and with the help of ${a_2}$ we can find ${b_3}$ and so on. In the last we get a series of $\cos$ functions. The applied unit on this sequence will get the required result.

Complete step by step solutions:
We have given that $a,b$ are positive quantities and $a$ is smaller than$b$. Also we have
${a_1} = \dfrac{{a + b}}{2}$ , ${b_1} = \sqrt {{a_1}{b_1}}$, ${a_2} = \dfrac{{a{}_1 + {b_1}}}{2}$ , ${b_2} = \sqrt {{a_2}{b_2}}$and so on…
Let us consider that $a = b\cos \theta$
We have given that ${a_1} = \dfrac{{a + b}}{2}$
So ${a_1} = \dfrac{{b\cos \theta + b}}{2}$
$\Rightarrow \dfrac{b}{2}\left( {\cos \theta + 1} \right)$
$\Rightarrow {a_1} = \dfrac{b}{2}(1 + \cos \theta )$
We know that 1 + \cos \theta = 2{\cos ^2}{\raise0.5ex\hbox{\scriptstyle 0} \kern-0.1em/\kern-0.15em \lower0.25ex\hbox{\scriptstyle 2}}
Therefore ${a_1} = \dfrac{b}{2}\left( {2{{\cos }^2}\dfrac{\theta }{2}} \right) = b{\cos ^2}\dfrac{\theta }{2}$
$\Rightarrow {a_1} = b{\cos ^2}\dfrac{\theta }{2}$
We have given that ${b_1} = \sqrt {{a_1}b}$
Putting value of ${a_1}$ , we get
${b_1} = \sqrt {b \times b{{\cos }^2}\dfrac{\theta }{2}} = \sqrt {{b^2}{{\cos }^2}{\raise0.5ex\hbox{$\scriptstyle \theta $} \kern-0.1em/\kern-0.15em \lower0.25ex\hbox{$\scriptstyle 2$}}} = b\cos {\raise0.5ex\hbox{$\scriptstyle \theta $} \kern-0.1em/\kern-0.15em \lower0.25ex\hbox{$\scriptstyle 2$}}$
So value of ${b_1} = b\cos \dfrac{\theta }{2}$
Also we have given that ${a_2} = \dfrac{{{a_1} + {b_1}}}{2}$putting the value of ${a_1}$ and ${a_2}$ in
${a_2} = \dfrac{{b{{\cos }^2}\dfrac{\theta }{2} + b\cos \dfrac{\theta }{2}}}{2} = \dfrac{{b\cos \dfrac{\theta }{2}\left( {\cos \dfrac{\theta }{2} + 1} \right)}}{2}$
$\Rightarrow \dfrac{{b\cos \dfrac{\theta }{2} + b{{\cos }^2}\dfrac{\theta }{2}}}{2} = b\cos \dfrac{\theta }{2}{\cos ^2}\dfrac{\theta }{4}$
${b_2} = \sqrt {{a_2}{b_1}}$
$\Rightarrow \sqrt {{b^2}{{\cos }^2}\dfrac{\theta }{2}{{\cos }^2}\dfrac{\theta }{4}} = b\cos \dfrac{\theta }{2}\cos \dfrac{\theta }{4}$
$\Rightarrow {b_2} = b\cos \dfrac{\theta }{2}\cos \dfrac{\theta }{4}$
So from the above sequence of ${b_2}$ we can write the value of ${b_3}$
${b_3} = b\cos \dfrac{\theta }{2}.\cos \dfrac{\theta }{4}.\cos \dfrac{\theta }{8}$
So, ${b_n} = b\cos \dfrac{\theta }{2}\cos \dfrac{\theta }{4}\cos \dfrac{\theta }{8}\cos \dfrac{\theta }{{16}},........\cos \dfrac{\theta }{{{2^n}}}$
Now ${b_\infty } = \mathop {\lim }\limits_{x \to \infty } bn$
Therefore ${b_\infty } = \mathop {\lim }\limits_{x \to \infty }$ $b\cos \dfrac{\theta }{2}\cos \dfrac{\theta }{4}\cos \dfrac{\theta }{8}.......\cos \dfrac{\theta }{{{2^n}}}$
Now we have $\sin \theta = 2\sin \dfrac{\theta }{2}\cos \dfrac{\theta }{2}$
Therefore $\cos \dfrac{\theta }{2} = \dfrac{{\sin \theta }}{{2\sin \dfrac{\theta }{2}}}$
We get ${b_\infty } = \mathop {\lim }\limits_{x \to \infty }$ $b\dfrac{{\sin \theta }}{{2\sin \dfrac{\theta }{2}}} \times \dfrac{{\sin \dfrac{\theta }{2}}}{{2\sin \dfrac{\theta }{4}}} \times \dfrac{{\sin \dfrac{\theta }{4}}}{{2\sin \dfrac{\theta }{8}}}$
${b_\infty } = \mathop {\lim }\limits_{x \to \infty } \dfrac{{b\sin \theta }}{{{2^n}\sin \dfrac{\theta }{{{2^n}}}}}$
Now ${b_\infty } = \mathop {\lim }\limits_{x \to \infty }$ $b\dfrac{\theta }{{{2^n}}} \times \dfrac{{\sin \theta }}{{\theta \sin \dfrac{\theta }{{{2^n}}}}}$
\Rightarrow \mathop {\lim }\limits_{x \to \infty } \dfrac{{b\sin \theta }}{\theta }.\left( {\dfrac{{{{{\raise0.5ex\hbox{\scriptstyle \theta } \kern-0.1em/\kern-0.15em \lower0.25ex\hbox{\scriptstyle 2}}}^n}}}{{\sin {{{\raise0.5ex\hbox{\scriptstyle \theta } \kern-0.1em/\kern-0.15em \lower0.25ex\hbox{\scriptstyle 2}}}^n}}}} \right)
$\Rightarrow b\dfrac{{\sin \theta }}{\theta }.\mathop {\lim }\limits_{x \to \infty } \left( {\dfrac{{\dfrac{\theta }{{{2^n}}}}}{{\sin {{{\raise0.5ex\hbox{$\scriptstyle \theta $} \kern-0.1em/\kern-0.15em \lower0.25ex\hbox{$\scriptstyle 2$}}}^n}}}} \right)$
$\Rightarrow \dfrac{{b\sin \theta }}{\theta } \times 1$
$\Rightarrow {b_\infty } = b\dfrac{{\sin \theta }}{\theta }$
Now $a = b\cos \theta$
$\Rightarrow \cos \theta = \dfrac{a}{b}$
$\sin \theta = \sqrt {1 - {{\cos }^2}\theta } = \sqrt {1 - \dfrac{{{a^2}}}{{{b^2}}}}$
Also $\cos \theta = \dfrac{\theta }{b}$
$\theta = {\cos ^{ - 1}}\dfrac{a}{b}$
So, ${b_\infty } = \dfrac{{b\sqrt {1 - \dfrac{{{a^2}}}{{{b^2}}}} }}{{{{\cos }^{ - 1}}\dfrac{a}{b}}} = \dfrac{{b\sqrt {\dfrac{{{b^2} - {a^2}}}{b}} }}{{{{\cos }^{ - 1}}\dfrac{a}{b}}}$
$\Rightarrow {b_\infty } = \dfrac{{\sqrt {{b^2} - {a^2}} }}{{{{\cos }^{ - 1}}\dfrac{a}{b}}}$
This is the requested result.

Therefore option $(B)$ is the correct answer.

Note:
Trigonometry is the branch of mathematics concerned with specific functions of angles and this application to calculations. There are six trigonometric functions of angles commonly used in trigonometry. Their names are $\sin e(\sin ),\cos ine(\cos ),$ $\tan gent(\tan ),\cot anget(\cot ),$ $\cot a\sec ant(\sec ),\cos ecant(\cos ec)$.
These trigonometric functions are related to the angle and the ratio of the sides of the triangle.