Question

(i) If $\cos x=-\frac{3}{5}$ and x lies in the IIIrd quadrant, find the values of
$\cos \frac{x}{2},\sin \frac{x}{2},\sin 2x$.

(ii) If $\cos x=-\frac{3}{5}$ and x lies in IInd quadrant, find the values of sin 2x and $\sin \frac{x}{2}$

Answer 1

(i) $\cos x=-\frac{3}{5}$

Using the identity $\cos 2\mathrm{\theta }={\cos }^2\mathrm{\theta }-{\sin }^2\mathrm{\theta }$, we get

$$\begin{gathered} \mathrm{cosx}={\cos }^2\frac{x}{2}-{\sin }^2\frac{x}{2} \\ \Rightarrow -\frac{3}{5}=2{\cos }^2\frac{x}{2}-1 \\ \Rightarrow 1-\frac{3}{5}=2{\cos }^2\frac{x}{2} \\ \Rightarrow \frac{2}{5}=2{\cos }^2\frac{x}{2} \\ \Rightarrow \frac{1}{5}={\cos }^2\frac{x}{2} \\ \Rightarrow \cos \frac{x}{2}=\pm \sqrt{\frac{1}{5}} \end{gathered}$$

It is given that x lies in the third quadrant. This means that $\frac{x}{2}$ lies in the second quadrant.

$\therefore \cos \frac{x}{2}=-\frac{1}{\sqrt{5}}$
Again,

$$\begin{gathered} \cos x={\cos }^2\frac{x}{2}-{\sin }^2\frac{x}{2} \\ \Rightarrow -\frac{3}{5}={\left(-\frac{1}{\sqrt{5}}\right)}^2-{\sin }^2\frac{x}{2} \\ \Rightarrow -\frac{3}{5}=\frac{1}{5}-{\sin }^2\frac{x}{2} \\ \Rightarrow -\frac{1}{5}-\frac{3}{5}=-{\sin }^2\frac{x}{2} \\ \Rightarrow \frac{4}{5}={\sin }^2\frac{x}{2} \\ \Rightarrow \sin \frac{x}{2}=\pm \frac{2}{\sqrt{5}} \end{gathered}$$

It is given that x lies in the third quadrant. This means that $\frac{x}{2}$ lies in the second quadrant.

$\therefore \sin \frac{x}{2}=\frac{2}{\sqrt{5}}$

$$\begin{gathered} \mathrm{Now}, \\ \sin x\mathit{}=\sqrt{1-{\cos }^{2}x} \\ \Rightarrow \sin x={\sqrt{1-\left(-\frac{3}{5}\right)}}^2 \\ \sin x=\sqrt{1-\frac{9}{25}}=\pm \frac{4}{5} \end{gathered}$$

Since x lies in the third quadrant, sinx is negative.

$$\begin{gathered} \therefore \sin x=-\frac{4}{5} \\ \Rightarrow \sin 2x=2\sin x\cos x \\ \Rightarrow \sin 2x=2\times \left(-\frac{4}{5}\right)\times \left(-\frac{3}{5}\right) \\ \Rightarrow \sin 2x=\frac{24}{25} \end{gathered}$$

(ii) $\cos x=-\frac{3}{5}$
$$\begin{gathered} \sin x=\sqrt{1-{\cos }^{2}x}=\sqrt{1-\left(\frac{-3}{5}\right)} \\ \Rightarrow \sin x=\pm \frac{4}{5} \end{gathered}$$
Here, x lies in the second quadrant.

$\therefore \sin x=\frac{4}{5}$

We know,

sin2x = 2sinx cosx

$\therefore \sin 2x=2\times \frac{4}{5}\times \left(-\frac{3}{5}\right)=-\frac{24}{25}$
Now,
$$\begin{gathered} \cos x=1-2{\sin }^2\frac{x}{2} \\ \Rightarrow 2{\sin }^2\frac{x}{2}=1-\left(-\frac{3}{5}\right)=\frac{8}{5} \\ \Rightarrow {\sin }^2\frac{x}{2}=\frac{4}{5} \\ \Rightarrow \sin \frac{x}{2}=\pm \frac{2}{\sqrt{5}} \end{gathered}$$

Since x lies in the second quadrant, $\frac{x}{2}$ lies in the first quadrant.
$\therefore \sin \frac{x}{2}=\frac{2}{\sqrt{5}}$