The value of 2 sin² B + 4 cos (A + B) sin A sin B + cos 2 (A + B) is

(a) 0
(b) cos 3 A
(c) cos 2 A
(d) None of these

Open in App

Solution

(c) cos 2A

$2 \sin^{2} B + 4 \cos (A + B) \sin A \sin B + \cos 2 (A + B) = 2 \sin^{2} B + 4 \cos A \cos B \sin A \sin B - 4 \sin^{2} A \sin^{2} B + \cos (A + B + A + B) = 2 \sin^{2} B + 4 \cos A \cos B \sin A \sin B - 4 \sin^{2} A \sin^{2} B + \cos^{2} (A + B) - \sin^{2} (A + B) [Using \cos (x + x) = \cos^{2} x - \sin^{2} x] = 2 \sin^{2} B + 4 \cos A \cos B \sin A \sin B - 4 \sin^{2} A \sin^{2} B + \cos^{2} A \cos^{2} B + \sin^{2} A \sin^{2} B - 2 \cos A \cos B \sin A \sin B - \sin^{2} A \cos^{2} B - \cos^{2} A \sin^{2} B - 2 \sin A \sin B \cos A \cos = \sin^{2} B + \sin^{2} B - 3 \sin^{2} A \sin^{2} B + \cos^{2} A \cos^{2} B - \sin^{2} A \cos^{2} B - \cos^{2} A \sin^{2} B = 1 - \cos^{2} B - 3 \sin^{2} A \sin^{2} B + \sin^{2} B + \cos^{2} A \cos^{2} B - \sin^{2} A \cos^{2} B - \cos^{2} A \sin^{2} B = 1 + \sin^{2} B (1 - \cos^{2} A) - 3 \sin^{2} A \sin^{2} B - \cos^{2} B (1 - \cos^{2} A) - \sin^{2} A \cos^{2} B = 1 + \sin^{2} B \sin^{2} A - 3 \sin^{2} A \sin^{2} B - \sin^{2} A \cos^{2} B - \cos^{2} B \sin^{2} A = 1 - 2 \sin^{2} A (\sin^{2} B + \cos^{2} B) = 1 - 2 \sin^{2} A = \cos^{2} A - \sin^{2} A = \cos 2 A$

Suggest Corrections

Similar questions

2 \sin^{2} B + 4 \cos (A + B) \sin A \sin B + \cos 2 (A + B)

The value of $s i n^{2}$ B+ 4 cos (A + B) sin B + cos 2(A + B) is

If a= $l o g_{12} 18$ , b= $l o g_{24} 54$ then the value of ab+5(a-b) is

[\vec{a} - \vec{b}, \vec{b} - \vec{c}, \vec{c} - \vec{a}], where |\vec{a}| = 1, |\vec{b}| = 5, |\vec{c}| = 3, is

2 \tan \frac{π}{10} + 3 \sec \frac{π}{10} - 4 \cos \frac{π}{10}

\sqrt{5}

Join BYJU'S Learning Program