Question

Let $ABC$ be a acute angled triangle such that $A=\frac{\pi }{3}$ and $\cos B\cos C=P$. The possible range of values of $P$ will be

• A. ($-\frac{1}{4}$,$\frac{1}{4}$]
• B. $(0,1]$
• C. $[\frac{1}{3},\infty )$
• D. $[1,\infty )$

Answer 1

The correct option is A ($-\frac{1}{4}$,$\frac{1}{4}$]

$\cos (B+C)=\cos B\cos C-\sin B\sin C$ ---------------(1)

and

$\cos (B-C)=\cos B\cos C+\sin B\sin C$ ----------------(2)

Adding (1) and (2)

$\cos (B+C)+\cos (B-C)=2\cos B\cos C=2P$

$\therefore P=\frac{\cos (B+C)+\cos (B-C)}{2}$--------------(3)

$A+B+C=\pi$ (Sum of angles of a triangle)

$B+C=\pi -A$

Taking cosine on both sides

$\cos (B+C)=\cos (\pi -A)=-\cos A$ (Using $\cos (\pi -x)=-\cos x$)

$\cos (B+C)=-\cos \frac{\pi }{3}=-\frac{1}{2}$

Substituting in (3),

$P=\frac{-\frac{1}{2}+\cos (B-C)}{2}$-------------------(4)

The range of values of cosine is $[-1,1]$.

This happens when the angle ranges from $\pi$ to $0$.

Substituting $1$ for $\cos (B-C)$ in (4),

$P=\frac{-\frac{1}{2}+1}{2}=\frac{1}{4}$.

This happens when $B-C=0$

$B=C$.

But $A=\frac{\pi }{3}$

$\therefore B=C=\frac{\pi }{3}$ Angles are less than $\frac{\pi }{2}$.

Hence acute angle criteria is satisfied.

Now consider the case where $\cos (B-C)=-1$

Then$P=\frac{-\frac{1}{2}+(-1)}{2}=-\frac{3}{4}$

But when $\cos (B-C)=-1,B-C=\pi$. This is not possible as both $B$ and $C$ should be acute angles.

Also because $B+C=\frac{2\pi }{3}$ , their difference cannot be greater than $\frac{2\pi }{3}$.

As these two angles are less than $\frac{\pi }{2}$, their difference will be less than $\frac{\pi }{2}$.

Hence range of $\cos (B-C)$ is $(0,1]$.

Value of $P$ at $\cos (A-B)=0$,

$P=\frac{-\frac{1}{2}+0}{2}=-\frac{1}{4}$

Hence range of $P$ is $(-\frac{1}{4},\frac{1}{4}]$