If-A-B-0-pi-2-such-that-3sin-2-A-2sin-2-B-1-and-3sin2A-2sin2B-0-find-the-value-of-A-B-

Question Number 27820 by bmind4860 last updated on 15/Jan/18

I f A, B ϵ (0, \frac{π}{2}) s u c h t h a t 3 {s i n}^{2} A + 2 {s i n}^{2} B = 1

a n d 3 s i n 2 A - 2 s i n 2 B = 0, f i n d t h e v a l u e o f A + B .

Answered by ajfour last updated on 15/Jan/18

\frac{3}{2} (1 - \cos 2 A) + 1 - \cos 2 B = 1

\Rightarrow 3 \cos 2 A + 2 \cos 2 B = 3 \dots . (i)

a n d s i n c e 3 \sin 2 A = 2 \sin 2 B

9 \sin^{2} 2 A = 4 \sin^{2} 2 B

\Rightarrow 9 - 9 \cos^{2} 2 A = 4 - 4 \cos^{2} 2 B

\Rightarrow 9 \cos^{2} 2 A - 4 \cos^{2} 2 B = 5

o r (3 \cos 2 A - 2 \cos 2 B) (3 \cos 2 A + 2 \cos 2 B) = 5

u s i n g (i) :

3 \cos 2 A - 2 \cos 2 B = \frac{5}{3}

a n d a s 3 \cos 2 A + 2 \cos 2 B = 3

a d d i n g t h e a b o v e t w o e q s .

6 \cos 2 A = \frac{14}{3} \Rightarrow \cos 2 A = \frac{7}{9}

a n d \sin 2 A = \sqrt{1 - \frac{49}{81}} = \frac{4 \sqrt{2}}{9}

s u b t r a c t i n g t h e m,

4 \cos 2 B = \frac{4}{3} \Rightarrow \cos 2 B = \frac{1}{3}

a n d \sin 2 B = \sqrt{1 - \frac{1}{9}} = \frac{2 \sqrt{2}}{3}

\cos (2 A + 2 B) = \cos 2 A \cos 2 B

- \sin 2 A \sin 2 B

= \frac{7}{9} \times \frac{1}{3} - \frac{4 \sqrt{2}}{9} \times \frac{2 \sqrt{2}}{3}

= - \frac{1}{3}

\cos (A + B) = \sqrt{\frac{1 + \cos (2 A + 2 B)}{2}}

= \frac{1}{\sqrt{3}}

A + B = \cos^{- 1} (\frac{1}{\sqrt{3}}) .

Commented by bmind4860 last updated on 15/Jan/18

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