Question

If in a $\Delta ABC$ , $2{{b}^{2}}={{a}^{2}}+{{c}^{2}}$ , then $\dfrac{\sin 3B}{\sin B}$ is equal to
(a).$\dfrac{{{c}^{2}}-{{a}^{2}}}{2ca}$
(b.$\dfrac{{{c}^{2}}-{{a}^{2}}}{ca}$
(c).${{\left( \dfrac{{{c}^{2}}-{{a}^{2}}}{ca} \right)}^{2}}$
(d).${{\left( \dfrac{{{c}^{2}}-{{a}^{2}}}{2ca} \right)}^{2}}$

Answer 1

Hint: For this question we will need formula for $\sin 3B$ in terms of $\sin B$ which is $\sin 3B=3\sin B-4{{\sin }^{3}}B$ and the relation $2{{b}^{2}}={{a}^{2}}+{{c}^{2}}$ hints us that the triangle is not right angled triangle and this relation may be useful in cosine rule for any general triangle which is $\cos A=\dfrac{{{b}^{2}}+{{c}^{2}}-{{a}^{2}}}{2bc}$ where ‘a’ is the side opposite to the angle A and ‘b’ and ‘c’ are the remaining two sides.

Complete step-by-step answer:
Now let us first evaluate $\dfrac{\sin 3B}{\sin B}$ ,for this we have the formula $\sin 3B=3\sin B-4{{\sin }^{3}}B$ . Therefore, substituting this value we have,
$\dfrac{\sin 3B}{\sin B}=\dfrac{3\sin B-4{{\sin }^{3}}B}{\sin B}$
Cancelling $\sin B$ from numerator and denominator we have,
$\dfrac{\sin 3B}{\sin B}=3-4{{\sin }^{2}}B$ …(i)

Now we need ${{\sin }^{2}}B$ from the information given in the question. But we do not have any direct relation between $\sin B$ and the terms given in the options but we do have a cosine rule which somewhat relates to our options. Now we can use the following relation to eliminate ${{\sin }^{2}}B$ and introduce $\cos B$ in our equation:
${{\sin }^{2}}B+{{\cos }^{2}}B=1\Rightarrow {{\sin }^{2}}B=1-{{\cos }^{2}}B$

Now let us substitute this in equation (i).
$\begin{align}
  & \dfrac{\sin 3B}{\sin B}=3-4{{\sin }^{2}}B \\
 & =3-4(1-{{\cos }^{2}}B) \\
\end{align}$

On further simplification we have,
$\dfrac{\sin 3B}{\sin B}=4{{\cos }^{2}}B-1$ …(ii)
Now let us find $\cos B$ using cosine rule,
$\cos B=\dfrac{{{a}^{2}}+{{c}^{2}}-{{b}^{2}}}{2ac}$

We are given that $2{{b}^{2}}={{a}^{2}}+{{c}^{2}}\Rightarrow {{b}^{2}}=\dfrac{{{a}^{2}}+{{c}^{2}}}{2}$ . Substituting this value in the above equation we get,
$\cos B=\dfrac{{{a}^{2}}+{{c}^{2}}-\left( \dfrac{{{a}^{2}}+{{c}^{2}}}{2} \right)}{2ac}$
On simplification we get,
$\cos B=\dfrac{{{a}^{2}}+{{c}^{2}}}{4ac}$

 Now let us substitute this value in equation (ii)
$\begin{align}
  & \dfrac{\sin 3B}{\sin B}=4{{\cos }^{2}}B-1 \\
 & =4{{\left( \dfrac{{{a}^{2}}+{{c}^{2}}}{4ac} \right)}^{2}}-1 \\
\end{align}$

Now we will evaluate further to make it similar to options in our question.
$\begin{align}
  & 4{{\left( \dfrac{{{a}^{2}}+{{c}^{2}}}{4ac} \right)}^{2}}-1=4\left( \dfrac{{{a}^{4}}+{{c}^{4}}+2{{a}^{2}}{{c}^{2}}}{16{{a}^{2}}{{c}^{2}}} \right)-1 \\
 & =\left( \dfrac{{{a}^{4}}+{{c}^{4}}+2{{a}^{2}}{{c}^{2}}}{4{{a}^{2}}{{c}^{2}}} \right)-1 \\
 & =\dfrac{{{a}^{4}}+{{c}^{4}}-2{{a}^{2}}{{c}^{2}}}{4{{a}^{2}}{{c}^{2}}} \\
\end{align}$

On completing square we get,
$\dfrac{\sin 3B}{\sin B}={{\left( \dfrac{{{c}^{2}}-{{a}^{2}}}{2ac} \right)}^{2}}$
Hence, option (d) is the correct option.

Note: For solving these kinds of questions we should thoroughly remember the formulas of trigonometry and properties of triangles and be intuitive about which formula can be used with the given amount of information.