If-sin-45-x-1-3-where-45-lt-x-lt-90-What-is-the-value-of-6sin-x-2-2-

Question Number 116441 by bobhans last updated on 04/Oct/20

If \sin (45 ° - x) = - \frac{1}{3}, where 45 ° < x < 90 °

. What is the value of {(6 \sin x - \sqrt{2})}^{2}

Answered by bemath last updated on 04/Oct/20

\Rightarrow \sin (x - 45) = \frac{1}{3}

\Rightarrow \frac{\sqrt{2}}{2} \sin x - \frac{\sqrt{2}}{2} \cos x = \frac{1}{3}

\Rightarrow \sqrt{2} \sin x - \sqrt{2} \cos x = \frac{2}{3}

\Rightarrow \sin x - \cos x = \frac{\sqrt{2}}{3}

\Rightarrow \sin x - \frac{\sqrt{2}}{3} = \sqrt{1 - \sin^{2} x}

\Rightarrow \sin^{2} x - \frac{2 \sqrt{2}}{3} \sin x + \frac{2}{9} = 1 - \sin^{2} x

\Rightarrow 2 \sin^{2} x - \frac{2 \sqrt{2}}{3} \sin x - \frac{7}{9} = 0

\Rightarrow \sin x = \frac{\frac{2 \sqrt{2}}{3} + \sqrt{\frac{8}{9} + \frac{56}{9}}}{4}

\Rightarrow \sin x = \frac{2 \sqrt{2} + 8}{12} = \frac{4 + \sqrt{2}}{6}

\Rightarrow 6 \sin x = 4 + \sqrt{2}

\Rightarrow {(6 \sin x - \sqrt{2})}^{2} = 16

Commented by bobhans last updated on 04/Oct/20

cooll

Answered by Dwaipayan Shikari last updated on 04/Oct/20

\frac{1}{\sqrt{2}} cosx - \frac{1}{\sqrt{2}} sinx = - \frac{1}{3}

sinx - cosx = \frac{\sqrt{2}}{3}

\sin^{2} x - 2 \frac{\sqrt{2}}{3} sinx + \frac{2}{9} = \cos^{2} x

{2 \sin}^{2} x - \frac{2 \sqrt{2}}{3} sinx - \frac{7}{9} = 0

sinx = \frac{\frac{2 \sqrt{2}}{3} \pm \sqrt{\frac{8}{9} + \frac{56}{9}}}{4} = \frac{\frac{\sqrt{2}}{3} \pm \frac{4}{3}}{2} = \frac{4 \pm \sqrt{2}}{6}

{(6 sinx - \sqrt{2})}^{2} = 4^{2} = 16 (As \frac{π}{4} < x < \frac{π}{2})