# 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 $\sin\frac{x}{2}$ .

#### 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
Is there an error in this question or solution?

#### 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