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प्रश्न
If \[\cos x = - \frac{3}{5}\] and x lies in the IIIrd quadrant, find the values of \[\cos\frac{x}{2}, \sin\frac{x}{2}, \sin 2x\] .
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उत्तर
\[ \Rightarrow - \frac{3}{5} = 2 \cos^2 \frac{x}{2} - 1\]
\[ \Rightarrow 1 - \frac{3}{5} = 2 \cos^2 \frac{x}{2}\]
\[ \Rightarrow \frac{2}{5} = 2 \cos^2 \frac{x}{2}\]
\[ \Rightarrow \frac{1}{5} = \cos^2 \frac{x}{2}\]
\[ \Rightarrow \cos\frac{x}{2} = \pm \sqrt{\frac{1}{5}}\]
\[ \Rightarrow - \frac{3}{5} = \left( - \frac{1}{\sqrt{5}} \right)^2 - \sin^2 \frac{x}{2}\]
\[ \Rightarrow - \frac{3}{5} = \frac{1}{5} - \sin^2 \frac{x}{2}\]
\[ \Rightarrow - \frac{1}{5} - \frac{3}{5} = - \sin^2 \frac{x}{2}\]
\[ \Rightarrow \frac{4}{5} = \sin^2 \frac{x}{2}\]
\[ \Rightarrow \sin\frac{x}{2} = \pm \frac{2}{\sqrt{5}}\]
\[\text{ sin } x = \sqrt{1 - \cos^2 x}\]
\[ \Rightarrow \text{ sin } x = \sqrt{1 - \left( - \frac{3}{5} \right)}^2 \]
\[\text{ sin } x = \sqrt{1 - \frac{9}{25}} = \pm \frac{4}{5}\]
\[ \Rightarrow \sin2x = 2\text{ sin } x\text{ cos } x\]
\[ \Rightarrow \sin2x = 2 \times \left( - \frac{4}{5} \right) \times \left( - \frac{3}{5} \right)\]
\[ \Rightarrow \sin2x = \frac{24}{25}\]
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