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Question Number 186144 by Rupesh123 last updated on 01/Feb/23

Commented by mr W last updated on 01/Feb/23

$i f t h e a g e o f y o u r f a t h e r i s 3 t i m e s o f$ $y o u r a g e, t h e n h o w f a r i s y o u r s c h o o l$ $a w a y f r o m y o u r h o m e ?$
Commented by Frix last updated on 01/Feb/23

$\frac{\cos 3 x}{\cos x} = \frac{1}{3} \Rightarrow \frac{\sin 3 x}{\sin x} = \frac{7}{3}$
Commented by mr W last updated on 01/Feb/23

$\frac{\cos 3 x}{\cos x} = \frac{1}{3} i s a n e q u a t i o n w h i c h$ $d e l i v e r s v a l u e s f o r x .$ $i n \int_{A}^{B} \frac{\sin 3 x}{\sin x} d x t h e \frac{\sin 3 x}{\sin x} i s a f u n c t i o n$ $o f x w h i c h i s i n t e g r a t e d f o r x f r o m$ $A t o B .$ $\int_{A}^{B} \frac{\sin 3 x}{\sin x} d x c a n a l s o b e w r i t t e n a s$ $\int_{A}^{B} \frac{\sin 3 t}{\sin t} d t o r \int_{A}^{B} \frac{\sin 3 u}{\sin u} d u e t c ., i t$ $d e l i v e r s a v a l u e w h i c h o n l y d e p e n d s o n$ $A a n d B, n o m a t t e r i f y o u w r i t e$ $\int_{A}^{B} \frac{\sin 3 x}{\sin x} d x o r \int_{A}^{B} \frac{\sin 3 t}{\sin t} d t o r \int_{A}^{B} \frac{\sin 3 u}{\sin u} d u .$ $t h e s o l u t i o n o f t h e f i r s t e q u a t i o n$ $h a s n o t h i n g t o d o w i t h t h e i n t e g r a l!$
Commented by Frix last updated on 01/Feb/23

$Yes of course you are right .$ $But as you certainly know with \frac{\sin 3 x}{\sin x} = \frac{7}{3}$ $we have \frac{7}{3} \underset{A}{\int^{B}} d x = \frac{7 (B - A)}{3}$
Commented by Frix last updated on 01/Feb/23

$. . . many strange questions here . . .$
Commented by mr W last updated on 01/Feb/23

$n o, i {d o n}^{'} t a g r e e w i t h y o u .$ $w h e n \int_{A}^{B} \frac{\sin 3 x}{\sin x} d x i s w r i t t e n, i t m e a n s$ $x i s a v a r i a b l e i n t h e r a n g e [A, B] .$ $t h e r e i s n o r e a s o n t o s a y$ $\frac{\sin 3 x}{\sin x} = c o n s t a n t = \frac{7}{3} .$ $y o u c o u l d b e r i g h t, w h e n i t i s$ $\int_{A}^{B} \frac{\sin 3 x}{\sin x} d t .$
Commented by ajfour last updated on 01/Feb/23

$I f \frac{\sin 3 x}{\sin x} = \frac{7}{3}$ $\Rightarrow 3 \sin 3 x = 7 \sin x$ $3 \sin x (3 - 4 \sin^{2} x) = 7 \sin x$ $e i t h e r \sin x = 0$ $o r 4 \sin^{2} x = 3 - \frac{7}{3} = \frac{2}{3}$ $\sin^{2} x = \frac{1}{6}$ $\sin x = \pm \frac{1}{\sqrt{6}}$ $s a y p = \sin^{- 1} \frac{1}{\sqrt{6}}$ $A = - p, B = p$ $(\frac{7}{3}) (B - A) = \frac{14}{3} \sin^{- 1} \frac{1}{\sqrt{6}}$
Commented by prakash jain last updated on 02/Feb/23

$\frac{\sin 3 x}{\sin x} = \frac{7}{3} \Rightarrow x is constant$ $\Rightarrow \int_{A}^{B} \frac{\sin 3 x}{\sin x} d x = 0$
Commented by ajfour last updated on 02/Feb/23

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