Given-sin-2a-sin-2b-cos-2a-cos-2b-2-3-find-the-value-of-cos-a-b-

Question Number 105290 by bemath last updated on 27/Jul/20

G i v e n \frac{\sin 2 a - \sin 2 b}{\cos 2 a + \cos 2 b} = \frac{2}{3}

f i n d t h e v a l u e o f \cos (a - b)

Answered by bobhans last updated on 27/Jul/20

\frac{2 \cos (\frac{2 a + 2 b}{2}) \sin (\frac{2 a - 2 b}{2})}{2 \cos (\frac{2 a + 2 b}{2}) \cos (\frac{2 a - 2 b}{2})} = \frac{2}{3}

\tan (a - b) = \frac{2}{3} \to {\begin{cases} \sin (a - b) = \frac{2}{\sqrt{13}} \\ \cos (a - b) = \frac{3}{\sqrt{13}} \end{cases}

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Commented by malwaan last updated on 27/Jul/20

t a n (a - b) = \frac{2}{3}

\frac{s i n (a - b)}{c o s (a - b)} = \frac{2}{3}

c o s (a - b) = \frac{3}{2} s i n (a - b)

= \frac{3}{2} \sqrt{1 - {c o s}^{2} (a - b)}

\Rightarrow \frac{4}{9} {c o s}^{2} (a - b) = 1 - {c o s}^{2} (a - b)

\Rightarrow \frac{13}{9} {c o s}^{2} (a - b) = 1

\Rightarrow c o s (a - b) = \pm \frac{3}{\sqrt{13}}

; s i n (a - b) = \frac{2}{3} c o s (a - b)

= \pm \frac{2}{\sqrt{13}}

Answered by Dwaipayan Shikari last updated on 27/Jul/20

\frac{2 c o s (a + b) s i n (a - b)}{2 c o s (a + b) c o s (a - b)} = \frac{2}{3}

\frac{\sqrt{1 - {c o s}^{2} (a - b)}}{c o s (a - b)} = \frac{2}{3}

\frac{1}{{c o s}^{2} (a - b)} = \frac{4}{9} + 1

{c o s}^{2} (a - b) = \frac{9}{13}

c o s (a - b) = \pm \frac{3}{\sqrt{13}}