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Question: Let $0 < A, B < \frac{\pi}{2}$ satisfying the equalities $3 \sin^2 A + 2 \sin^2 B = 1$ and $3\sin2A ...

Let 0<A,B<π20 < A, B < \frac{\pi}{2} satisfying the equalities 3sin⁡2A+2sin⁡2B=13 \sin^2 A + 2 \sin^2 B = 1 and 3sin⁡2A−2sin⁡2B=03\sin2A - 2\sin2B = 0. Then A+2B=A + 2B = :

A

π4\frac{\pi}{4}

B

π3\frac{\pi}{3}

C

π2\frac{\pi}{2}

D

None of these.

Answer

π2\frac{\pi}{2}

Explanation

Solution

The given equations are:

  1. 3sin⁡2A+2sin⁡2B=13 \sin^2 A + 2 \sin^2 B = 1
  2. 3sin⁡2A−2sin⁡2B=03 \sin 2A - 2 \sin 2B = 0

From (2), 3sin⁡2A=2sin⁡2B3 \sin 2A = 2 \sin 2B.

Using sin⁡2x=1−cos⁡2x2\sin^2 x = \frac{1 - \cos 2x}{2}, equation (1) becomes: 3(1−cos⁡2A2)+2(1−cos⁡2B2)=13 \left(\frac{1 - \cos 2A}{2}\right) + 2 \left(\frac{1 - \cos 2B}{2}\right) = 1 32−32cos⁡2A+1−cos⁡2B=1\frac{3}{2} - \frac{3}{2} \cos 2A + 1 - \cos 2B = 1 52−32cos⁡2A−cos⁡2B=1\frac{5}{2} - \frac{3}{2} \cos 2A - \cos 2B = 1 32=32cos⁡2A+cos⁡2B\frac{3}{2} = \frac{3}{2} \cos 2A + \cos 2B 3=3cos⁡2A+2cos⁡2B3 = 3 \cos 2A + 2 \cos 2B (Equation 1')

From (2), 3sin⁡2A=2sin⁡2B3 \sin 2A = 2 \sin 2B. Squaring both sides: 9sin⁡22A=4sin⁡22B9 \sin^2 2A = 4 \sin^2 2B 9(1−cos⁡22A)=4(1−cos⁡22B)9 (1 - \cos^2 2A) = 4 (1 - \cos^2 2B) 9−9cos⁡22A=4−4cos⁡22B9 - 9 \cos^2 2A = 4 - 4 \cos^2 2B 5=9cos⁡22A−4cos⁡22B5 = 9 \cos^2 2A - 4 \cos^2 2B (Equation 2')

Let's test the option A+2B=π2A + 2B = \frac{\pi}{2}. This implies A=π2−2BA = \frac{\pi}{2} - 2B. Then 2A=π−4B2A = \pi - 4B. cos⁡2A=cos⁡(π−4B)=−cos⁡4B=−(2cos⁡22B−1)=1−2cos⁡22B\cos 2A = \cos(\pi - 4B) = -\cos 4B = -(2\cos^2 2B - 1) = 1 - 2\cos^2 2B. sin⁡2A=sin⁡(π−4B)=sin⁡4B=2sin⁡2Bcos⁡2B\sin 2A = \sin(\pi - 4B) = \sin 4B = 2\sin 2B \cos 2B.

Substitute into 3sin⁡2A=2sin⁡2B3 \sin 2A = 2 \sin 2B: 3(2sin⁡2Bcos⁡2B)=2sin⁡2B3 (2\sin 2B \cos 2B) = 2 \sin 2B Since 0<B<π20 < B < \frac{\pi}{2}, sin⁡2B≠0\sin 2B \neq 0. 6cos⁡2B=2  ⟹  cos⁡2B=136 \cos 2B = 2 \implies \cos 2B = \frac{1}{3}.

Now find cos⁡2A\cos 2A using cos⁡2B=13\cos 2B = \frac{1}{3}: cos⁡2A=1−2cos⁡22B=1−2(13)2=1−2(19)=1−29=79\cos 2A = 1 - 2\cos^2 2B = 1 - 2(\frac{1}{3})^2 = 1 - 2(\frac{1}{9}) = 1 - \frac{2}{9} = \frac{7}{9}.

Check with equation (1'): 3cos⁡2A+2cos⁡2B=3(79)+2(13)=73+23=93=33 \cos 2A + 2 \cos 2B = 3(\frac{7}{9}) + 2(\frac{1}{3}) = \frac{7}{3} + \frac{2}{3} = \frac{9}{3} = 3. This is consistent. The ranges 0<A,B<π20 < A, B < \frac{\pi}{2} are also satisfied.