If-angles-A-and-B-satisfy-2-cos-A-cos-B-cos-3-B-and-2-sin-A-sin-B-sin-3-B-then-the-value-of-1620sin-2-A-B-is-

Question Number 18891 by Tinkutara last updated on 01/Aug/17

If angles A and B satisfy \sqrt{2} \cos A =

\cos B + \cos^{3} B and \sqrt{2} \sin A = \sin B -

\sin^{3} B, then the value of {1620 \sin}^{2} (A - B)

is

Answered by behi.8.3.4.1.7@gmail.com last updated on 02/Aug/17

2 {c o s}^{2} A = {c o s}^{2} B {(1 + {c o s}^{2} B)}^{2}

2 {s i n}^{2} A = {s i n}^{2} B {(1 - {s i n}^{2} B)}^{2}

c o s B = t

\Rightarrow 2 = t^{2} {(1 + t^{2})}^{2} + (1 - t^{2}) . t^{4} = t^{2} (1 + 3 t^{2})

\Rightarrow 3 t^{4} + t^{2} - 2 = 0 \Rightarrow t^{2} = \frac{- 1 + \sqrt{1 + 24}}{6} = \frac{2}{3}

\Rightarrow {c o s}^{2} B = \frac{2}{3}, {s i n}^{2} B = \frac{1}{3}

\Rightarrow {c o s}^{2} A = \frac{1}{3} . {(1 + \frac{2}{3})}^{2} = \frac{1}{3} . \frac{25}{9} = \frac{25}{27}

\Rightarrow {c o s}^{2} A = \frac{25}{27}, {s i n}^{2} A = \frac{2}{27}

s i n (A - B) = s i n A . c o s B - c o s A . s i n B =

= \frac{\sqrt{2}}{3 \sqrt{3}} . \frac{\sqrt{2}}{\sqrt{3}} - \frac{5}{3 \sqrt{3}} . \frac{1}{\sqrt{3}} = \frac{2}{9} - \frac{5}{9} = - \frac{1}{3}

\Rightarrow 1620 \times {s i n}^{2} (A - B) = 1620 \times \frac{1}{9} = 180 .

Answered by ajfour last updated on 01/Aug/17

\sqrt{2} \sin (A - B) = \sqrt{2} \sin Acos B - \sqrt{2} \cos Asin B

= (\sin B - \sin^{3} B) \cos B

- (\cos B + \cos^{3} B) \sin B

= - (\sin Bcos B) (1)

\Rightarrow 2 \sin^{2} (A - B) = \sin^{2} Bcos^{2} B . . (i)

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

and \sqrt{2} \sin A = \sin B (1 - \sin^{2} B)

squaring and adding above two eqns :

2 = \cos^{2} B {(1 + \cos^{2} B)}^{2} + (1 - \cos^{2} B) {(\cos^{2} B)}^{2}

2 = \cos^{2} B + 2 \cos^{4} B + \cos^{6} B

+ \cos^{4} B - \cos^{6} B

\Rightarrow 3 \cos^{4} B + \cos^{2} B - 2 = 0

3 \cos^{4} B + 3 \cos^{2} B - 2 \cos^{2} B - 2 = 0

\Rightarrow (3 \cos^{2} B - 2) (\cos^{2} B + 1) = 0

\Rightarrow \cos^{2} B = \frac{2}{3} and \sin^{2} B = \frac{1}{3}

Using in (i) :

2 \sin^{2} (A - B) = \sin^{2} Bcos^{2} B

= \frac{1}{3} \times \frac{2}{3} = \frac{2}{9}

So 1620 \sin^{2} (A - B) = 810 \times \frac{2}{9}

1620 \sin^{2} (A - B) = 180 .

Commented by behi.8.3.4.1.7@gmail.com last updated on 02/Aug/17

i t i s p e r f e c t m r A j f o u r .

Commented by ajfour last updated on 01/Aug/17

Is this correct ?

Commented by Tinkutara last updated on 01/Aug/17

Thank you very much ajfour Sir!

Commented by ajfour last updated on 01/Aug/17

thanks for confirming .

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