Question

Given that $\cos \frac{x}{2}=\frac{12}{13}$ and $x$ lies in first quadrant, calculate without the use of tables, the values of $sinx,cosx$ and $tanx$.

Answer 1

$\begin{array}{c}\cos \frac{x}{2}=\frac{12}{13}\\ 1+\cos x=2{\cos }^2\frac{x}{2}\\ \cos x=2{\cos }^2\frac{x}{2}-1=2\times {\left(\frac{12}{13}\right)}^2-1=2\times \left(\frac{144}{169}\right)-1=\frac{119}{169}\\ \left[\cos x=\frac{119}{169}\right]\\ Now,\\ si{n}^2\frac{x}{2}=1-{\cos }^2\frac{x}{2}=1-{\left(\frac{12}{13}\right)}^2=\frac{169-144}{169}=\frac{25}{169}\\ \sin \frac{x}{2}=\frac{5}{13}\\ \sin x=2\sin \frac{x}{2}\cos \frac{x}{2}=2\times \frac{5}{13}\times \frac{12}{13}=\frac{120}{169}\\ \left[\sin x=\frac{120}{169}\right]\\ \tan x=\frac{\cos x}{\sin x}=\frac{\frac{119}{169}}{\frac{120}{169}}=\frac{119}{120}\\ \left[\tan x=\frac{119}{120}\right]\end{array}$