Question

If A and B are positive acute angles satisfying the equalities3cos2A+ cos2B=4
and 3sinAsinB=2cosBcosA, thenA+2B is equal to
(A) 774
(B) T3
(C) TT6
(D) T2
A
B
ООО
С
D​

Answer 1

Given: A and B are positive acute angles satisfying the equalities

3 cos^2 A+ 2cos^2 B = 4  and 3sinA / sinB = 2cosB / cosA

To find: The value of A + 2B?

Solution:

• Now we have given the following equalities as:

                   3 cos^2 A + 2cos^2 B = 4

                   2cos^2 B - 1 = 4 - 3 cos^2 A - 1

                   cos 2B = 3  - 3 cos^2 A

                   cos 2B = 3 ( 1 - cos^2 A)

                   cos 2B = 3 sin^2 A     .......................(i)

• Now we have another inequality as:

                   3sinA / sinB = 2cosB / cosA

                   sin 2B = 3 sin A cos A  ......................(ii)

• Now we know that

                   cos (A + 2B) = cos A cos 2B - sin A sin 2B

• Now putting (i) and (ii) in above equation, we get:

                   cos (A + 2B) = cos A(3 sin^2 A) - sin A( 3 sin A cos A)

                   cos (A + 2B) = 0

                   A + 2B = π/2

Answer:

              So the value of A + 2B is π/2.