Here, x lies in 2nd quadrant.
Therefore,
\( \dfrac{\pi}{4}< \dfrac{x}{2}< \dfrac{\pi}{2}\)
hence
\( \dfrac{x}{2}\) lies in 1st quadrant.
Therefore,
\( sin\dfrac{x}{2},cos\dfrac{x}{2}\;and\;tan\dfrac{x}{2}\) are positive
Given that sinx = \( \dfrac{1}{4} \)
Now we know that, sin2x+cos2x=1
Therefore substituting \( si nx=\dfrac{1}{4}\) and computing ,
\( cosx=-\dfrac{\sqrt{15}}{4}\)
since x lies in 2nd quadrant, cosx is negative.
Now we know that,\( 2sin^2\dfrac{x}{2} =1-cosx\)
Computing we get, \( 2sin^2\dfrac{x}{2} =1+\dfrac{\sqrt{15}}{4}\)
Hence
\( sin\dfrac{x}{2}=\sqrt{\dfrac{4+\sqrt{15}}{8}}\)
Now we know that, sin2x+cos2x=1
Therefore substituting
\( sin\dfrac{x}{2}=\sqrt{\dfrac{4+\sqrt{15}}{8}}\) and computing,
\( cos\dfrac{x}{2}=\sqrt{\dfrac{4+\sqrt{15}}{8}}\)
Hence,
\(tan\dfrac{x}{2}=\dfrac{sinx\dfrac{x}{2}}{cos\dfrac{x}{2}}\)
\(=\dfrac{\sqrt{4+\sqrt{15}}}{\sqrt{4-\sqrt{15}}}\)
= \( 4\pm\sqrt{15}\)
Answered by Abhisek | 1 year agoprove that \(sin \dfrac{8π}{3} cos \dfrac{23π}{6} + cos \dfrac{13π}{3} sin \dfrac{35π}{6} = \dfrac{1}{2}\)
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