Question
Evaluate: $\frac{\cos 3 A-2 \cos 4 A}{\sin 3 A+2 \sin 4 A}$ , when $A = 15^\circ$
✓
Answer
Given that $A= 15^\circ$
$\frac{\cos 3 A -2 \cos 4 A }{\sin 3 A +2 \sin 4 A }$
$=\frac{\cos 3\left(15^{\circ}\right)-2 \cos 4\left(15^{\circ}\right)}{\sin 3\left(15^{\circ}\right)+2 \sin 4\left(15^{\circ}\right)}$
$=\frac{\cos 45^{\circ}-2 \cos 60^{\circ}}{\sin 45^{\circ}+2 \sin 60^{\circ}}$
$=\frac{\frac{1}{\sqrt{2}}-2\left(\frac{1}{2}\right)}{\frac{1}{\sqrt{2}}+2\left(\frac{\sqrt{3}}{2}\right)}$
$=\frac{\frac{1}{\sqrt{2}}-1}{\frac{1}{\sqrt{2}}+\sqrt{3}}$
$=\frac{\frac{1-\sqrt{2}}{\sqrt{2}}}{\frac{1+\sqrt{6}}{\sqrt{2}}}$
$=\frac{1-\sqrt{2}}{1+\sqrt{6}}$
$\frac{\cos 3 A -2 \cos 4 A }{\sin 3 A +2 \sin 4 A }$
$=\frac{\cos 3\left(15^{\circ}\right)-2 \cos 4\left(15^{\circ}\right)}{\sin 3\left(15^{\circ}\right)+2 \sin 4\left(15^{\circ}\right)}$
$=\frac{\cos 45^{\circ}-2 \cos 60^{\circ}}{\sin 45^{\circ}+2 \sin 60^{\circ}}$
$=\frac{\frac{1}{\sqrt{2}}-2\left(\frac{1}{2}\right)}{\frac{1}{\sqrt{2}}+2\left(\frac{\sqrt{3}}{2}\right)}$
$=\frac{\frac{1}{\sqrt{2}}-1}{\frac{1}{\sqrt{2}}+\sqrt{3}}$
$=\frac{\frac{1-\sqrt{2}}{\sqrt{2}}}{\frac{1+\sqrt{6}}{\sqrt{2}}}$
$=\frac{1-\sqrt{2}}{1+\sqrt{6}}$
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