Question

Show by the mathematical induction that
$\frac{1}{sin2x}+\frac{1}{sin4x}+\frac{1}{sin{2}^{n}x}=cotx-cot{2}^{n}x$

Answer 1

cot A - cot B = $\frac{sin(B-A)}{sinAsinB}$For p (1) L. H. S. = $\frac{1}{sin2x}$

R. H. S. = cot x - cot 2x = $\frac{sin(2x-x)}{sinxsin2x}=\frac{1}{sin2x}$

Thus p(1) is holds goods assume p(n) is true for all

p(n + 1) = p(n) + $\frac{1}{sin{2}^{n+1}x}$

= p(n) + $\frac{sin{2}^{n}x}{sin{2}^{n}xsin{2}^{n+1}x}$= p(n) + $\frac{sin(2^{n+1}-2^n)x}{sin{2}^{n}xsin{2}^{n+1}x}$

= $(cotx-cot{2}^{n}x)+(cot{2}^n-cot{2}^{n+1}x)$

$=cotx-cot{2}^{n+1}x$

Thus the p(n + 1) alos goods holds .