If 0 ≤ x ≤ π and x lies in the IInd quadrant such that sin x

Question:

(i) If $0 \leq x \leq \pi$ and $x$ lies in the llnd quadrant such that $\sin x=\frac{1}{4}$. Find the values of $\cos \frac{x}{2}, \sin \frac{x}{2}$ and $\tan \frac{x}{2}$

(ii) If $\cos x=\frac{4}{5}$ and $x$ is acute, find $\tan 2 x$

(iii) If $\sin x=\frac{4}{5}$ and $0

Solution:

(i) $\sin x=\frac{1}{4}$

$\therefore \sin x=\sqrt{1-\cos ^{2} x}$

$\Rightarrow\left(\frac{1}{4}\right)^{2}=1-\cos ^{2} x$

$\Rightarrow \frac{1}{16}-1=-\cos ^{2} x$

$\Rightarrow \frac{15}{16}=\cos ^{2} x$

$\Rightarrow \cos x=\pm \frac{\sqrt{15}}{4}$

Since x lies in the 2nd quadrant, cos x is negative.

Thus,

$\cos x=-\frac{\sqrt{15}}{4}$

Now, using the identity $\cos x=2 \cos ^{2} \frac{x}{2}-1$, we get

$-\frac{\sqrt{15}}{4}=2 \cos ^{2} \frac{x}{2}-1$

$\Rightarrow-\frac{\sqrt{15}}{8}=\cos ^{2} \frac{x}{2}-\frac{1}{2}$

$\Rightarrow \cos ^{2} \frac{x}{2}=\frac{4-\sqrt{15}}{8}$

 

$\Rightarrow \cos \frac{x}{2}=\pm \frac{4-\sqrt{15}}{8}$

Since $x$ lies in the 2 nd quadrant and $\frac{x}{2}$ lies in the 1 st quadrant, $\cos \frac{x}{2}$ is positive.

$\therefore \cos \frac{x}{2}=\frac{4-\sqrt{15}}{8}$

Again,

$\cos x=\cos ^{2} \frac{x}{2}-\sin ^{2} \frac{x}{2}$

$\Rightarrow-\frac{\sqrt{15}}{4}=\left(\sqrt{\frac{4-\sqrt{15}}{8}}\right)^{2}-\sin ^{2} \frac{x}{2}$

 

$\Rightarrow-\frac{\sqrt{15}}{4}=\frac{4-\sqrt{15}}{8}-\sin ^{2} \frac{x}{2}$

$\Rightarrow \sin ^{2} \frac{x}{2}=\frac{4+\sqrt{15}}{8}$

 

$\Rightarrow \sin \frac{x}{2}=\pm \sqrt{\frac{4+\sqrt{15}}{8}}=\sqrt{\frac{4+\sqrt{15}}{8}}$

Now,

$\tan \frac{x}{2}=\frac{\sin \frac{x}{2}}{\cos \frac{x}{2}}$

$=\frac{\sqrt{\frac{4+\sqrt{15}}{8}}}{\sqrt{\frac{4-\sqrt{15}}{8}}}=\sqrt{\frac{4+\sqrt{15}}{4-\sqrt{15}}}$

$=\sqrt{\frac{(4+\sqrt{15})(4+\sqrt{15})}{(4-\sqrt{15})(4+\sqrt{15})}}$

 

$=\frac{4+\sqrt{15}}{4^{2}-(\sqrt{15})^{2}}=\frac{4+\sqrt{15}}{16-15}=4+\sqrt{15}$

(ii) $\cos x=\frac{4}{5}$

$\therefore \sin x=\sqrt{1-\cos ^{2} x}$

$=\sqrt{1-\left(\frac{4}{5}\right)^{2}}$

 

$=\sqrt{1-\frac{16}{25}}$

$=\sqrt{\frac{25-16}{25}}$

$=\sqrt{\frac{9}{25}}$

 

$=\frac{3}{5}$

$\therefore \tan x=\frac{\sin x}{\cos x}$

$=\frac{\frac{3}{5}}{\frac{4}{5}}$

 

$=\frac{3}{4}$

Now,

$\tan 2 x=\frac{2 \tan x}{1-\tan ^{2} x}$

$=\frac{2\left(\frac{3}{4}\right)}{1-\left(\frac{3}{4}\right)^{2}}$

 

$=\frac{2\left(\frac{3}{4}\right)}{1-\frac{9}{16}}$

$=\frac{\frac{3}{2}}{\frac{7}{16}}$

 

$=\frac{24}{7}$

Hence, the value of $\tan 2 x$ is $\frac{24}{7}$.

(iii) $\sin x=\frac{4}{5}$ and $0

$\therefore \sin x=\sqrt{1-\cos ^{2} x}$

$\Rightarrow\left(\frac{4}{5}\right)^{2}=1-\cos ^{2} x$

 

$\Rightarrow \frac{16}{25}-1=-\cos ^{2} x$

$\Rightarrow \frac{9}{25}=\cos ^{2} x$

 

$\Rightarrow \cos x=\pm \frac{3}{5}$

Since x lies in the 1st quadrant, cos x is positive.

Thus,

$\cos x=\frac{3}{5}$

Now,

$\sin (4 x)=2 \sin (2 x) \cos (2 x)$

$=2(2 \sin x \cos x)\left(1-2 \sin ^{2} x\right)$

 

$=2\left(2 \times \frac{4}{5} \times \frac{3}{5}\right)\left(1-2\left(\frac{4}{5}\right)^{2}\right)$

$=2\left(\frac{24}{25}\right)\left(1-\frac{32}{25}\right)$

 

$=2\left(\frac{24}{25}\right)\left(\frac{25-32}{25}\right)$

$=2\left(\frac{24}{25}\right)\left(\frac{-7}{25}\right)$

 

$=-\frac{336}{625}$

Hence, the value of $\sin 4 x$ is $-\frac{336}{625}$.

 

 

 

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