Question

(i) If $\cos \mathrm{\theta }=\frac{4}{5}$ and θ is acute, find tan 2θ

(ii) If $\sin \mathrm{\theta }=\frac{4}{5}$ and $0<\mathrm{\theta }<\frac{\pi }{2}$, find the value of sin 4 θ

Answer 1

(i) Given:
$\cos \mathrm{\theta }=\frac{4}{5}$
Thus,
$$\begin{gathered} \mathrm{tan\theta }=\frac{\mathrm{sin\theta }}{\mathrm{cos\theta }}=\frac{\sqrt{1-{\cos }^2\mathrm{\theta }}}{\mathrm{cos\theta }}=\frac{\sqrt{1-{\left({\displaystyle \frac{4}{5}}\right)}^2}}{{\displaystyle \frac{4}{5}}} \\ =\frac{\sqrt{1-\left({\displaystyle \frac{16}{25}}\right)}}{{\displaystyle \frac{4}{5}}}=\frac{3}{4}\left(\mathrm{Because}\mathrm{\theta }\mathrm{lies}\mathrm{in}\mathrm{the}{1}^{\mathrm{st}}\mathrm{quadrant},\mathrm{tan\theta }\mathrm{will}\mathrm{be}\mathrm{positive}.\right) \\ \mathrm{Now}, \\ \tan 2\mathrm{\theta }=\frac{2\times {\displaystyle \frac{3}{4}}}{1-{\left({\displaystyle \frac{3}{4}}\right)}^2}\left[\because \tan 2\mathrm{\theta }=\frac{2\mathrm{tan\theta }}{1-{\tan }^2\mathrm{\theta }}\right] \\ =\frac{{\displaystyle \frac{3}{2}}}{1-\left({\displaystyle \frac{9}{16}}\right)}=\frac{24}{7} \end{gathered}$$

(ii) Given:
$\sin \mathrm{\theta }=\frac{4}{5}$ and $0<\mathrm{\theta }<\frac{\pi }{2}$

Thus,

$$\begin{gathered} \mathrm{cos\theta }=\sqrt{1-{\sin }^2\mathrm{\theta }}=\sqrt{1-{\left(\frac{4}{5}\right)}^2}=\frac{3}{5}\left(\because 0<\mathrm{\theta }<\frac{\mathrm{\pi }}{2}\right) \\ \mathrm{We}\mathrm{know}, \\ \sin 2\mathrm{\theta }=2\mathrm{sin\theta cos\theta } \\ \therefore \sin 2\mathrm{\theta }=2\times \frac{4}{5}\times \frac{3}{5}=\frac{24}{25} \end{gathered}$$

$$\begin{gathered} \mathrm{Again}, \\ \cos 2\mathrm{\theta }=\sqrt{1-{\sin }^{2}2\mathrm{\theta }}=\sqrt{1-{\left(\frac{24}{25}\right)}^2}=\pm \frac{7}{25} \\ \mathrm{Because}\mathrm{\theta }\mathrm{lies}\mathrm{in}\mathrm{the}{1}^{\mathrm{st}}\mathrm{quadrant}\mathrm{and}2\mathrm{\theta }\mathrm{lies}\mathrm{in}\mathrm{the}{2}^{\mathrm{nd}}\mathrm{quadrant},\cos 2\mathrm{\theta }\mathrm{will}\mathrm{be}\mathrm{negative}. \\ \therefore \cos 2\mathrm{\theta }=-\frac{7}{25} \\ \mathrm{Now}, \\ \mathrm{We}\mathrm{know},\sin 4\mathrm{\theta }=2\sin 2\mathrm{\theta cos}2\mathrm{\theta } \\ \therefore \sin 4\mathrm{\theta }=2\times \frac{24}{25}\times \left(-\frac{7}{25}\right)=-\frac{336}{625} \end{gathered}$$