Question

Evaluate
sin 11x/12​

Answer 1

$\textbf{To find:}$

$\text{The value of sin\frac{11\,\pi}{12} }$

$\textbf{Solution:}$

$\text{We know that,}$

$\boxed{\bf\,cos2A=1-2\,sin^2A}$

$\implies\,sin^2A=\dfrac{1-cos\,2A}{2}$

$\text{Put A=\dfrac{11\pi}{12} , we get}$

$\implies\,sin^2\frac{11\pi}{12}=\dfrac{1-cos\,2(\frac{11\pi}{12})}{2}$

$\implies\,sin^2\frac{11\pi}{12}=\dfrac{1-cos\frac{11\pi}{6}}{2}$

$\implies\,sin^2\frac{11\pi}{12}=\dfrac{1-cos\,330^{\circ}}{2}$

$\implies\,sin^2\frac{11\pi}{12}=\dfrac{1-sin\,60^{\circ}}{2}$

$\implies\,sin^2\frac{11\pi}{12}=\dfrac{1-\frac{\sqrt3}{2}}{2}$

$\implies\,sin^2\frac{11\pi}{12}=\dfrac{\frac{2-\sqrt3}{2}}{2}$

$\implies\,sin^2\frac{11\pi}{12}=\dfrac{2-\sqrt3}{4}$

$\implies\,sin\dfrac{11\pi}{12}=\pm\dfrac{\sqrt{2-\sqrt3}}{2}$

$\text{But \frac{11\pi}{12} lies in II quadrant. sin\frac{11\pi}{12} is positive}$

$\therefore\boxed{\bf\,sin\frac{11\pi}{12}=\dfrac{\sqrt{2-\sqrt3}}{2}}$

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