Trigonometrical Ratios And Identities Ques 45
Prove that the values of the function $\frac{\sin x \cos 3 x}{\sin 3 x \cos x}$ do not lie between $\frac{1}{3}$ and $3$ for any real $x$.
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Solution:
Formula:
- Let $y=\frac{\sin x \cos 3 x}{\sin 3 x \cos x}=\frac{\tan x}{\tan 3 x}$
$\Rightarrow \quad y=\frac{\tan x}{\tan 3 x}=\frac{\tan x\left(1-3 \tan ^{2} x\right)}{3 \tan x-\tan ^{3} x} $
$=\frac{1-3 \tan ^{2} x}{3-\tan ^{2} x}$
Put $\tan x=t$
$\Rightarrow y=\frac{1-3 t^{2}}{3-t^{2}} $
$\Rightarrow 3 y-t^{2} y=1-3 t^{2} $
$\Rightarrow 3 y-1=t^{2} y-3 t^{2} $
$\Rightarrow 3 y-1=t^{2}(y-3) $
$\Rightarrow \frac{3 y-1}{y-3}=t^{2} \Rightarrow \frac{3 y-1}{y-3}>0 $
$\therefore \quad t^{2}>0 $
NOTE: It is a brilliant technique to convert equation into inequation and asked in IIT papers frequently.
$\quad \Rightarrow y<1 / 3$ or $y>3$.
This shows that $y$ cannot lie between $1 / 3$ and 3 .
BETA
