IIT-JEE 1997Question 1
Prove that the values of the function $\dfrac{\sin x\cos3x}{\sin3x\cos x}$ do not lie between $\dfrac13$ and $3$ for any real $x$.
Trigonometric Ratio & Identities
Answerproved
Solution
Concept. The function is $\frac{\tan x}{\tan3x}$, a function of $u=\tan^2x\ge0$; solving for $u$ shows which values are possible. Formulas used. (1) $\tan3x=\dfrac{3t-t^3}{1-3t^2}$ with $t=\tan x$Step 1 - Rewrite. By (1), with $u=t^2$: $$y=\frac{\tan x}{\tan3x}=\frac{t(1-3t^2)}{3t-t^3}=\frac{1-3u}{3-u}.$$ Step 2 - Solve for $u$. $y(3-u)=1-3u$ gives $$u=\frac{1-3y}{3-y}.$$ Step 3 - Sign. If $\frac13<y<3$, then $1-3y<0$ and $3-y>0$, so $u<0$. But $u=\tan^2x\ge0$. So no such $y$ occurs. Hence proved.