Question

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

(ii) If $cos\theta =\frac{4}{5}and\theta$ is acute, find $tan2\theta$

(iii) If $\theta =\frac{4}{5}and0<\theta <\frac{\pi }{2}$, find the value of sin $4\theta$.

Answer 1

(i) Since x lies in $I{I}^{nd}$ quadrant

$\Rightarrow \frac{\pi }{2}<x<\pi$
$\Rightarrow \frac{\pi }{4}<\frac{x}{2}<\frac{\pi }{2},$ which means $\frac{x}{2}$ lies in $1^{st}$ quad.

Now,

$sinx=\frac{1}{4}=\frac{p}{h}\Rightarrow p=1\Rightarrow b=\sqrt{15}$

So, $cosx=\frac{b}{h}=\frac{-\sqrt{15}}{4}$

(-ve due to $I{I}^{nd}$ quad)

Thus,

$cos\frac{x}{2}=\sqrt{\frac{1+cosx}{2}}=\sqrt{\frac{1-\frac{\sqrt{15}}{4}}{2}}=\frac{\sqrt{4-\sqrt{15}}}{8}sin\frac{x}{2}=\sqrt{\frac{1-cosx}{2}}=\sqrt{\frac{1+\frac{\sqrt{15}}{4}}{2}}=\frac{\sqrt{4-\sqrt{15}}}{8}$

$tan\frac{x}{2}=\frac{sin\frac{x}{2}}{cos\frac{x}{2}}=\frac{\sqrt{\frac{4+\sqrt{15}}{8}}}{\sqrt{\frac{4-\sqrt{15}}{8}}}=\sqrt{\frac{4+\sqrt{15}}{4-\sqrt{15}}}=\sqrt{\frac{(1+\sqrt{15})(4+\sqrt{15})}{(4-\sqrt{15})(4+\sqrt{15})}}=4+\sqrt{15}$

(ii) Since $\theta$ in acute, so $0\le 2\theta <\pi$

Now, $\cos \theta =\frac{4}{5}=\frac{b}{h}$
$\Rightarrow b=4\Rightarrow p=3$

h = 5

$\therefore sin\theta =\frac{p}{h}=\frac{3}{5}tan\theta =\frac{p}{b}=\frac{3}{4}\text{so},tan2\theta =\frac{2tan\theta }{1-ta{n}^2\theta }=\frac{2.\frac{3}{4}}{1-{\left(\frac{3}{4}\right)}^2}=\frac{\frac{6}{4}}{\frac{7}{16}}=\frac{24}{7}$

(iii) $sin\theta =\frac{4}{5}=\frac{p}{h}\Rightarrow p=4$

$\Rightarrow$ b = 3

h = 5

$\therefore cos\theta =\frac{b}{h}=\frac{3}{5}$

Now, $sin\theta .cos\theta =2.\frac{4}{5}.\frac{3}{5}=\frac{24}{25}$

$cos2\theta =co{s}^2\theta -si{n}^2\theta ={\left(\frac{3}{5}\right)}^2-{\left(\frac{4}{5}\right)}^2=\frac{-7}{25}\text{So},sin4\theta =sin2.2\theta =2sin2\theta .cos2\theta =2.\frac{24}{25}.\left(\frac{-7}{25}\right)=\frac{-336}{625}$