If sinA+cos2A =(1/2) and cosA+sin2A=(1/3) , then find the value of sin3A.


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Question Number 51069 by rahul 19 last updated on 23/Dec/18

$I f \sin A + \cos 2 A = \frac{1}{2} a n d$ $\cos A + \sin 2 A = \frac{1}{3}, t h e n f i n d t h e v a l u e$ $o f \sin 3 A .$
	Answered by peter frank last updated on 23/Dec/18

	$\cos 2 A = \frac{1}{2} - \sin A . . . . . (i)$ $\sin 2 A = \frac{1}{3} - \cos A . . . . . (i i)$ $s q u a r e (i) a n d (i i) a n d a d d$ $1 = \frac{1}{4} + \frac{1}{9} - \sin A - \frac{2}{3} \cos A + 1$ $\sin A - \frac{2}{3} \cos A = - \frac{13}{36}$ $3 \sin A - 2 \cos A = - \frac{13}{9}$ ${(3 \sin A - 2 \cos A)}^{3} = {(- \frac{13}{9})}^{3}$ $39 \sin^{3} A - 50 \sin^{2} A \cos A + 12 \sin A - 8 \cos A = - \frac{2197}{729} . . . . . (i i i)$ $\sin 2 A = \frac{1}{3} - \cos A . . . . . (i i)$ $c o s A = \frac{1}{3 + 6 s i n A} . . . . . (i v)$ $p l e a s e w a i t . . . .$
	Answered by MJS last updated on 23/Dec/18

	$\sin A + \cos 2 A = \frac{1}{2} \Leftrightarrow \sin^{2} A - \frac{1}{2} \sin A - \frac{1}{4} = 0 \Rightarrow$ $\Rightarrow \sin A = \frac{1}{4} \pm \frac{\sqrt{5}}{4} (1)$ $\cos A + \sin 2 A = \frac{1}{3} \Leftrightarrow 2 \sin A \cos A + \cos A - \frac{1}{3} = 0 \Rightarrow$ $\Rightarrow \cos A = \frac{1}{3 (1 + 2 \sin A)} (2)$ $(1) in (2)$ $\cos A = \frac{1}{2} \pm \frac{\sqrt{5}}{6}$ $but (\sin A = \frac{1}{4} \pm \frac{\sqrt{5}}{4}) \land (\cos A = \sqrt{1 - \sin^{2} A}) \Rightarrow$ $\Rightarrow \cos A = \frac{1}{4} \sqrt{10 \mp 2 \sqrt{5}} \neq \frac{1}{2} \pm \frac{\sqrt{5}}{6} \Rightarrow$ $\Rightarrow no solution for A$
	Commented by rahul 19 last updated on 24/Dec/18

	$T h a n k y o u s i r!$ $B u t w h a t w e n t w r o n g i n T a n m a y {S i r}^{'} s$ $s o l u t i o n ?$
	Commented by mr W last updated on 24/Dec/18

	$M J S s i r i s r i g h t . t h e v a l u e f o r$ $\sin A + \cos 2 A a n d t h e v a l u e f o r$ $\cos A + \sin 2 A c a n n o t b e c h o s e n$ $i n d e p e n d e n t l y f r o m e a c h o t h e r . i f$ $\sin A + \cos 2 A = \frac{1}{2}, t h e n$ $\cos A + \sin 2 A c a n o n l y b e \pm \frac{\sqrt{5 + 2 \sqrt{5}}}{2}$ $o r \pm \frac{\sqrt{5 - 2 \sqrt{5}}}{2}, b u t n e v e r \frac{1}{3} .$
	Commented by MJS last updated on 24/Dec/18

	$you can ‘ ‘ {solve}^{″} in many ways$ $\sin^{2} A - \frac{1}{2} \sin A - \frac{1}{4} = 0$ $2 \sin A \cos A + \cos A - \frac{1}{3} = 0$ $write s for \sin A and c for \cos A$ $s^{2} - \frac{1}{2} s - \frac{1}{4} = 0 (1)$ $2 s c + c - \frac{1}{3} = 0 (2)$ $we can solve this system for s and c . but s and$ $c {aren}^{'} t independent . we have a third equation$ $s^{2} + c^{2} = 1 (3)$ $\Rightarrow no solution$
	Commented by rahul 19 last updated on 24/Dec/18

	$T h a n k y o u b o t h s i r s!$ $U n d e r s t o o d n o w .$
	Answered by mr W last updated on 24/Dec/18

	$t h e q u e s t i o n s h o u l d h a v e b e e n :$ $i f \sin A + \cos 2 A = a a n d$ $\cos A + \sin 2 A = b, t h e n f i n d t h e v a l u e$ $o f \sin 3 A .$ $s o l u t i o n :$ $\sin A + \cos 2 A = a$ $\sin^{2} A + 2 \sin A \cos 2 A + \cos^{2} 2 A = a^{2}$ $\cos A + \sin 2 A = b$ $\cos^{2} A + 2 \cos A \sin 2 A + \sin^{2} 2 A = b^{2}$ $\Rightarrow 1 + 2 (\sin A \cos 2 A + \cos A \sin 2 A) + 1 = a^{2} + b^{2}$ $\Rightarrow 1 + 2 \sin 3 A + 1 = a^{2} + b^{2}$ $\Rightarrow \sin 3 A = \frac{a^{2} + b^{2}}{2} - 1$ $n o t e : a a n d b c a n n o t b e a n y v a l u e s!$ $t h e y m u s t m a t c h t o e a c h o t h e r . e . g .$ $a = \frac{1}{2} a n d b = \frac{1}{3} {d o n}^{'} t m a t c h . i f a = \frac{1}{2},$ $t h e n b c a n o n l y b e \pm \frac{\sqrt{5 + 2 \sqrt{5}}}{2} o r \pm \frac{\sqrt{5 - 2 \sqrt{5}}}{2} .$ $e x a m p l e :$ $a = \frac{1}{2}, b = \frac{\sqrt{5 + 2 \sqrt{5}}}{2}$ $\Rightarrow \sin 3 A = \frac{1}{2} (\frac{1}{4} + \frac{5 + 2 \sqrt{5}}{4}) - 1 = \frac{\sqrt{5} - 1}{4}$
	Commented by rahul 19 last updated on 24/Dec/18

	$T h a n k y o u S i r!$
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