Question

$K=\frac{cos3A-2cos4A}{sin3A+2sin4A}$, when A = ${15}^{\circ }$,$K=\frac{1-\sqrt{2}}{1+\sqrt{6}}$

write 1 for true and 0 for false

Answer 1

Given, $A={15}^{\circ }$
Thus, $\frac{\cos 3A-2\cos 4A}{\sin 3A+2\sin 4A}$
Put $A={15}^{\circ }$
= $\frac{\cos 45-2\cos 60}{\sin 45+2\sin 60}$
= $\frac{\frac{1}{\sqrt{2}}-2\left(\frac{1}{2}\right)}{\frac{1}{\sqrt{2}}+2(\frac{\sqrt{3}}{2}}$
= $\frac{\frac{1}{\sqrt{2}}-1}{\frac{1}{\sqrt{2}}+\sqrt{3}}$
= $\frac{\frac{1-\sqrt{2}}{\sqrt{2}}}{\frac{1}{\sqrt{2}}+\sqrt{3}}$
= $\frac{1-\sqrt{2}}{1+\sqrt{6}}$