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If Cos X = − 3 5 and X Lies in Iind Quadrant, Find the Values of Sin 2x and Sin X 2 . - Mathematics

Numerical

 If  \[\cos x = - \frac{3}{5}\]  and x lies in IInd quadrant, find the values of sin 2x and \[\sin\frac{x}{2}\] .

 

 

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Solution

\[\cos x = - \frac{3}{5}\]
\[\text{ sin } x = \sqrt{1 - \cos^2 x} = \sqrt{1 - \left( \frac{- 3}{5} \right)}\]
\[ \Rightarrow \text{ sin } x = \pm \frac{4}{5}\]

Here, x lies in the second quadrant.

\[\therefore \text{ sin } x = \frac{4}{5}\]
We know,
sin2x = 2sinx cosx
\[\therefore \sin2x = 2 \times \frac{4}{5} \times \left( - \frac{3}{5} \right) = - \frac{24}{25}\]
Now,
\[\text{ cos } x = 1 - 2 \sin^2 \frac{x}{2}\]
\[ \Rightarrow 2 \sin^2 \frac{x}{2} = 1 - \left( - \frac{3}{5} \right) = \frac{8}{5}\]
\[ \Rightarrow \sin^2 \frac{x}{2} = \frac{4}{5}\]
\[ \Rightarrow \sin\frac{x}{2} = \pm \frac{2}{\sqrt{5}}\]
Since x lies in the second quadrant,
\[\frac{x}{2}\]  lies in the first quadrant.
\[\therefore \sin\frac{x}{2} = \frac{2}{\sqrt{5}}\]
 
Concept: Values of Trigonometric Functions at Multiples and Submultiples of an Angle
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APPEARS IN

RD Sharma Class 11 Mathematics Textbook
Chapter 9 Values of Trigonometric function at multiples and submultiples of an angle
Exercise 9.1 | Q 28.2 | Page 29
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