If 2cos⁡A=cos⁡B+cos3⁡B, and 2sin⁡A=sin⁡B−sin3⁡B then sin (A- B)=

2cos⁡A=cos⁡B+cos3⁡B----iand 2sin⁡A=sin⁡B−sin3⁡B----ii⇒ 2sin⁡Acos⁡B−2cos⁡Asin⁡B =sin⁡B−sin3⁡Bcos⁡B−cos⁡B+cos3⁡Bsin⁡B =−sin⁡Bcos⁡B⇒ sin⁡(A−B)=−122sin⁡2BNow squaring and adding Eqs. (i) and (ii), we get 2=cos2⁡B+sin2⁡B+cos6⁡B+sin6⁡B+2cos4⁡B−sin2⁡B⇒1=cos2⁡A+sin2⁡A3−3cos2⁡Asin2⁡Acos2⁡A+sin2⁡A+2 cos 2B⇒1=1−(3/4)sin2⁡2B+2cos⁡2B⇒−3sin2⁡2B+8cos⁡2B=0⇒3cos2⁡2B+8cos⁡2B−3=0⇒cos⁡2B=13⇒ sin⁡2B=±223 ∴sin⁡(A−B)=±13

If 2cos⁡A=cos⁡B+cos3⁡B, and 2sin⁡A=sin⁡B−sin3⁡B then sin (A- B)=

2cos⁡A=cos⁡B+cos3⁡B----iand 2sin⁡A=sin⁡B−sin3⁡B----ii⇒ 2sin⁡Acos⁡B−2cos⁡Asin⁡B =sin⁡B−sin3⁡Bcos⁡B−cos⁡B+cos3⁡Bsin⁡B =−sin⁡Bcos⁡B⇒ sin⁡(A−B)=−122sin⁡2BNow squaring and adding Eqs. (i) and (ii), we get 2=cos2⁡B+sin2⁡B+cos6⁡B+sin6⁡B+2cos4⁡B−sin2⁡B⇒1=cos2⁡A+sin2⁡A3−3cos2⁡Asin2⁡Acos2⁡A+sin2⁡A+2 cos 2B⇒1=1−(3/4)sin2⁡2B+2cos⁡2B⇒−3sin2⁡2B+8cos⁡2B=0⇒3cos2⁡2B+8cos⁡2B−3=0⇒cos⁡2B=13⇒ sin⁡2B=±223 ∴sin⁡(A−B)=±13

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If $\sqrt{2} \cos A = \cos B + \cos^{3} B, and \sqrt{2} \sin A = \sin B - \sin^{3} B$ then sin (A- B)=

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$\pm 1$

$\pm \frac{1}{2}$

$\pm \frac{1}{3}$

$\pm \frac{1}{4}$

answer is C.

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Detailed Solution

$\sqrt{2} \cos A = \cos B + \cos^{3} B - - - - (i)$

and $\sqrt{2} \sin A = \sin B - \sin^{3} B - - - - (ii)$

$\begin{array}{l} \Rightarrow \sqrt{2} \sin Acos B - \sqrt{2} \cos Asin B \\ = (\sin B - \sin^{3} B) \cos B - (\cos B + \cos^{3} B) \sin B \\ = - \sin Bcos B \\ \Rightarrow \sin (A - B) = - \frac{1}{2 \sqrt{2}} \sin 2 B \end{array}$

Now squaring and adding Eqs. (i) and (ii), we get

$\begin{array}{l} 2 = \cos^{2} B + \sin^{2} B + \cos^{6} B + \sin^{6} B + 2 (\cos^{4} B - \sin^{2} B) \\ \Rightarrow 1 = {(\cos^{2} A + \sin^{2} A)}^{3} - 3 \cos^{2} {Asin}^{2} A (\cos^{2} A + \sin^{2} A) + 2 \cos 2 B \\ \Rightarrow 1 = 1 - (3 / 4) \sin^{2} 2 B + 2 \cos 2 B \\ \Rightarrow - 3 \sin^{2} 2 B + 8 \cos 2 B = 0 \\ \Rightarrow 3 \cos^{2} 2 B + 8 \cos 2 B - 3 = 0 \\ \Rightarrow \cos 2 B = \frac{1}{3} \\ \Rightarrow \sin 2 B = \pm \frac{2 \sqrt{2}}{3} ∴ \sin (A - B) = \pm \frac{1}{3} \end{array}$