Question

show that 16 sin^5theeta= sin5theeta-5sin3theeta+10sin theeta​

Answer 1

Answer:

$\mathrm{solve\:for\:a,\:16\sin ^5\left(t\right)heeta=\sin \left(5t\right)heeta-5\sin \left(3t\right)heeta+10\sin \left(t\right)heeta\quad :\quad a=0}$

Step-by-step explanation:

$\mathrm{16\sin ^5\left(t\right)heeta=\sin \left(5t\right)heeta-5\sin \left(3t\right)heeta+10\sin \left(t\right)heeta}$

$\mathrm{16\sin ^5\left(t\right)heeta=e^2hat\sin \left(5t\right)-5e^2hat\sin \left(3t\right)+10e^2hat\sin \left(t\right)}$

$\mathrm{16e^2hat\sin ^5\left(t\right)=e^2hat\sin \left(5t\right)-5e^2hat\sin \left(3t\right)+10e^2hat\sin \left(t\right)}$

$\mathrm{Subtract\:}e^2hat\sin \left(5t\right)-5e^2hat\sin \left(3t\right)+10e^2hat\sin \left(t\right)\mathrm{\:from\:both\:sides}$

$\mathrm{16e^2hat\sin ^5\left(t\right)-\left(e^2hat\sin \left(5t\right)-5e^2hat\sin \left(3t\right)+10e^2hat\sin \left(t\right)\right)=e^2hat\sin \left(5t\right)}$

$\mathrm{-5e^2hat\sin \left(3t\right) + 10e^2 hatsin(t) - (e^2 hatsin 5(t) - 5e^2hatsin(3t) + 10e^2hatsin(t))}$

$\mathrm{16e^2hat\sin ^5\left(t\right)-e^2hat\sin \left(5t\right)+5e^2hat\sin \left(3t\right)-10e^2hat\sin \left(t\right)=0}$

$\mathrm{e^2hat\left(16\sin ^5\left(t\right)-\sin \left(5t\right)+5\sin \left(3t\right)-10\sin \left(t\right)\right)=0}$

$\mathrm{Using\:the\:Zero\:Factor\:Principle:\quad \:If}\:ab=0\:\mathrm{then}\:a=0\:\mathrm{or}\:b=0$

$a=0$