Question

prove that cos cube pi by 9 + sin cube pi by 18 is equal to 3 by 4 into cos pi by 9 + sin pi by 18​

Answer 1

Answer:

$\cos^3{\frac{\pi}{9}}+\sin^3{\frac{\pi}{18}}$

Using

$\boxed{\begin{minipage}{6cm} \bf\:cos3A=4cos^3A-3cosA\\ \\\:\implies\bf{cos^3A=\frac{1}{4}[cos3A+3\:cosA]}\\ \\ sin3A=3\:sinA-4sin^3A\\ \\\:\implies\bf{sin^3A=\frac{1}{4}[3\:sinA-sin3A]} \end{minipage}}$

$=\frac{1}{4}[cos3\frac{\pi}{9}+3\:cos\frac{\pi}{9}]+\frac{1}{4}[3\:sin\frac{\pi}{18}-sin3\frac{\pi}{18}]$

$=\frac{1}{4}[cos\frac{\pi}{3}+3\:cos\frac{\pi}{9}]+\frac{1}{4}[3\:sin\frac{\pi}{18}-sin\frac{\pi}{6}]$

$=\frac{1}{4}[\frac{1}{2}+3\:cos\frac{\pi}{9}]+\frac{1}{4}[3\:sin\frac{\pi}{18}-\frac{1}{2}]$

$=\frac{1}{4}[\frac{1}{2}+3\:cos\frac{\pi}{9}+3\:sin\frac{\pi}{18}-\frac{1}{2}]$

$=\frac{1}{4}[3\:cos\frac{\pi}{9}+3\:sin\frac{\pi}{18}]$

$=\frac{3}{4}[cos\frac{\pi}{9}+sin\frac{\pi}{18}]$

$\implies\boxed{\cos^3{\frac{\pi}{9}}+\sin^3{\frac{\pi}{18}}=\frac{3}{4}[cos\frac{\pi}{9}+sin\frac{\pi}{18}]}$