Question

If ${\prod }_{r=4}^8\cos \left(\frac{\theta }{2^r}\right)=\frac{\sin \left(\frac{\theta }{2^{n_1}}\right)}{(2{)}^{n_2}\sin \left(\frac{\theta }{2^{n_3}}\right)},$ then the value of $n_1+n_3-n_2$ is

Answer 1

${\prod }_{r=4}^8\cos \left(\frac{\theta }{2^{}}\right)=\cos \left(\frac{\theta }{2^4}\right)\cos \left(\frac{\theta }{2^5}\right)\cos \left(\frac{\theta }{2^6}\right)\cos \left(\frac{\theta }{2^7}\right)\cos \left(\frac{\theta }{2^8}\right)$

$=\frac{2\cos \left(\frac{\theta }{2^8}\right)\sin \left(\frac{\theta }{2^8}\right)\cos \left(\frac{\theta }{2^7}\right)\cos \left(\frac{\theta }{2^6}\right)\cos \left(\frac{\theta }{2^5}\right)\cos \left(\frac{\theta }{2^4}\right)}{2\sin \left(\frac{\theta }{2^8}\right)}$

$=\frac{2.\sin \left(\frac{\theta }{2^{}}\right)\cos \left(\frac{\theta }{2^7}\right)\cos \left(\frac{\theta }{2^6}\right)\cos \left(\frac{\theta }{2^5}\right)\cos \left(\frac{\theta }{2^4}\right)}{2^2\sin \left(\frac{\theta }{2^8}\right)}$

$=\frac{2.\sin \left(\frac{\theta }{2^6}\right)\cos \left(\frac{\theta }{2^6}\right)\cos \left(\frac{\theta }{2^5}\right)\cos \left(\frac{\theta }{2^4}\right)}{2^3\sin \left(\frac{\theta }{2^8}\right)}=\frac{2.\sin \left(\frac{\theta }{2^5}\right)\cos \left(\frac{\theta }{2^5}\right)\cos \left(\frac{\theta }{2^4}\right)}{2^4\sin \left(\frac{\theta }{2^8}\right)}$

$=\frac{2.\sin \left(\frac{\theta }{2^4}\right)\cos \left(\frac{\theta }{2^4}\right)}{2^5\sin \left(\frac{\theta }{2^8}\right)}=\frac{2.\sin \left(\frac{\theta }{2^3}\right)}{2^5\sin \left(\frac{\theta }{2^8}\right)}=\frac{2.\sin \left(\frac{\theta }{2^{n_1}}\right)}{2^{n_2}\sin \left(\frac{\theta }{2^{n_3}}\right)}$

$n_1+n_3-n_2=3+8-5=11-5=6$