Here, x is 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 tanx = \(- \dfrac{4}{3} \)
We know that,
\( sec^2x=1+tan^2x\)
\(=1+(-\dfrac{4}{3})^2\)
= \(\dfrac{25}{9}\)
As x is in 2nd quadrant, secx is negative.
Therefore ,
\( secx=-\dfrac{5}{3}\)
Then
\( cosx=-\dfrac{3}{5}\)
Now we know that, \( 2cos^2\dfrac{x}{2}=cosx+1\)
Computing we get,
\(2cos^2 \dfrac{x}{2}= \dfrac{2}{5}\)
\( cos\dfrac{x}{2}=\dfrac{1}{\sqrt{5}}\)
Now we know that, sin2x+cos2x=1
Therefore substituting \( cos\dfrac{x}{2}=\dfrac{1}{\sqrt{5}}\) and computing,
\( sin\dfrac{x}{2}=\dfrac{2}{\sqrt{5}}\)
Hence ,
\( tan=\dfrac{x}{2}=\dfrac{sin\dfrac{x}{2}}{cos\dfrac{x}{2}}\) = 2
Thus, the respective values of \( sin\dfrac{x}{2},cos\dfrac{x}{2}\;and\;tan\dfrac{x}{2}\) are
\( \dfrac{2\sqrt{5}}{5}, \dfrac{\sqrt{5}}{5},2\)
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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