Question

If $15si{n}^{4}a+10co{s}^{4}a=6$, find the value of $8cose{c}^{6}a-27se{c}^{6}a$.

Answer 1

$\frac{140000}{15}si{n}^{4}a+10co{s}^{4}a=6$
$15si{n}^{4}a+10co{s}^{4}a=6(si{n}^{2}a+co{s}^{2}a{)}^2$
$15si{n}^{4}a+10co{s}^{4}a=6si{n}^{4}a+6co{s}^{4}a+12si{n}^{a}acos$
$9si{n}^{4}a+4co{s}^{4}a-12si{n}^{2}aco{s}^{2}a=0$
$(3si{n}^{2}a-2co{s}^{2}a{)}^2=0$
$\Rightarrow ta{n}^{2}a=\frac{2}{3}$
$So,se{c}^{2}a=1+ta{n}^{2}a=1+\frac{2}{3}=\frac{5}{3}$
$se{c}^{6}a={\left(\frac{5}{3}\right)}^3-\left(1\right)$
$Also,co{s}^{2}a=\frac{3}{5}\Rightarrow si{n}^{2}a=1-\frac{3}{5}$
$\therefore cose{c}^{2}a=\frac{5}{2}\Rightarrow cose{c}^{6}a={\left(\frac{5}{2}\right)}^3$
$\therefore 8cose{c}^{6}a-27se{c}^{6}a=8\times {\left(\frac{5}{2}\right)}^3-27\left(\frac{5}{3}\right)$
$=5^3-5^3$
= 0