Geometric Progression

Algebra Level 3

If 0 < x < π 2 0 < x < \frac{\pi}{2} and tan ⁡ x 2 = t , \tan \frac{x}{2}=t, what is the sum of the infinite geometric progression 1 + sin ⁡ x + sin ⁡ 2 x + sin ⁡ 3 + ⋯ ? 1+\sin x+\sin^2 x+\sin ^3 +\cdots?

( 1 − t ) 2 1 + t 2 \frac{(1-t)^2}{1+t^2} 1 + t 2 ( 1 + t ) 2 \frac{1+t^2}{(1+t)^2} ( 1 + t ) 2 1 + t 2 \frac{(1+t)^2}{1+t^2} 1 + t 2 ( 1 − t ) 2 \frac{1+t^2}{(1-t)^2}

View wiki
Mahindra Jain by Mahindra Jain

This section requires Javascript.
You are seeing this because something didn't load right. We suggest you, (a) try refreshing the page, (b) enabling javascript if it is disabled on your browser and, finally, (c) loading the non-javascript version of this page . We're sorry about the hassle.

1 solution

Tom Engelsman
Nov 8, 2020

The above infinite series sums to Σ k = 0 ∞ ( sin ⁡ x ) k = 1 1 − sin ⁡ x \Sigma_{k=0}^{\infty} (\sin x)^{k} = \frac{1}{1-\sin x} (i). If tan ⁡ x 2 = t \tan \frac{x}{2} = t , then x = 2 arctan ⁡ ( t ) x = 2\arctan(t) (ii). Substitution of (ii) into (i) now yields:

1 1 − sin ⁡ ( 2 arctan ⁡ ( t ) ) ; \frac{1}{1-\sin(2\arctan(t))} ;

or 1 1 − 2 sin ⁡ ( arcsin ⁡ ( t / t 2 + 1 ) ) cos ⁡ ( arccos ⁡ ( 1 / t 2 + 1 ) ) ; \frac{1}{1 - 2\sin(\arcsin(t / \sqrt{t^2+1})) \cos(\arccos(1 / \sqrt{t^2+1})) };

or 1 1 − 2 ( t / t 2 + 1 ) \frac{1}{1 - 2(t / t^2+1)} ;

or t 2 + 1 t 2 − 2 t + 1 ; \frac{t^2+1}{t^2 - 2t + 1};

or t 2 + 1 ( t − 1 ) 2 . \boxed{\frac{t^2+1}{(t-1)^2}}.

0 Helpful 0 Interesting 0 Brilliant 0 Confused

0 pending reports

×

Problem Loading...

Note Loading...

Set Loading...