Question

if ${\sin }^{3}x\sin 3x=\sum _{m=0}^6{c}_m{\cos }^{m}x,where{c}_{0,}{c}_{1,}.....c_6\text{constants,}\text{then:}$

• A. $c_0+c_2+c_4+c_6=0$
• B. $c_1+c_3+c_5=6$
• C. $2c_2+3c_6+=0$
• D. $c_4+2c_6+=0$

Answer 1

The correct option is A $c_0+c_2+c_4+c_6=0$

Given > ${\sin }^{3}x\sin 3x=\sum _{m=0}^6{c}_m{\cos }^{m}x$

${\sin }^{3}x\sin 3x={\sin }^{3}x[3\sin x-4{\sin }^{3}x]$

$=3{\sin }^{4}x-4{\sin }^{6}x$

$=3({\sin }^{2}x{)}^2-4({\sin }^{2}x{)}^3$

$=3(1-{\cos }^{2}x{)}^2-4(1-{\cos }^{2}x{)}^3$

$=-1+6{\cos }^{2}x-9{\cos }^{4}x+4{\cos }^{6}x$

$=C_0+C_1\cos x+C_2{\sin }^{2}x+C_3{\sin }^{3}x+C_4{\sin }^{4}x+C_5{\sin }^{5}x+C_6{\sin }^{6}x$

$\therefore C_0+C_2+C_4+C_6=-1+6-9+4=0$