Question bab8a

Aug 16, 2016

If $\phi$ is the angle of 3rd quadrant then

$\cos \phi \to - v e \mathmr{and} \sin \phi \to - v e$

Now $\sin 2 \phi = 2 \sin \phi \cos \phi$

$= - 2 \sqrt{1 - {\cos}^{2} \phi} \cdot \cos \phi$

$= - 2 \sqrt{1 - {\left(- \frac{15}{17}\right)}^{2}} \cdot \left(- \frac{15}{17}\right)$

$= + 2 \frac{\sqrt{{17}^{2} - {15}^{2}}}{17} \cdot \times \frac{15}{17}$

$= \frac{2 \cdot 8 \cdot 15}{17} ^ 2 = \frac{240}{289}$

$2 \phi = {\sin}^{-} 1 \left(\frac{240}{289}\right)$
$\approx {56.14}^{\circ} \textcolor{red}{\to 2 \phi \text{ is in the 1st quadrant}}$

Again

$\cos \phi = - \frac{15}{17}$

$\phi = {\cos}^{-} 1 \left(- \frac{15}{17}\right) = 360 - 151.93 = 208.07$
$\textcolor{red}{\left(\text{As "phi" is in the 3rd quadrant}\right)}$

:.2phi=416.14=360+56 .14#

So $2 \phi$ is the angle of 1st quadrant.