Evaluating the Series at \( x = \frac{\pi}{4} \)

The question asks for the value of the infinite series $1+ \sin x + \cos^2 x + \sin^3 x + \dots$ when $x = \frac{\pi}{4}$.

Substituting the Value of \( x \)

First, let's find the values of $\sin x$ and $\cos x$ at $x = \frac{\pi}{4}$:

• $ \sin\left(\frac{\pi}{4}\right) = \frac{1}{\sqrt{2}} $
• $ \cos\left(\frac{\pi}{4}\right) = \frac{1}{\sqrt{2}} $

Notice that at $x = \frac{\pi}{4}$, we have $ \sin x = \cos x = \frac{1}{\sqrt{2}} $. Let's substitute this value into the terms of the series:

• Term 1: $1$
• Term 2: $ \sin x = \frac{1}{\sqrt{2}} $
• Term 3: $ \cos^2 x = \left(\frac{1}{\sqrt{2}}\right)^2 = \frac{1}{2} $
• Term 4: $ \sin^3 x = \left(\frac{1}{\sqrt{2}}\right)^3 = \frac{1}{2\sqrt{2}} $

If we assume the pattern continues by alternating $\sin$ and $\cos$ powers, the subsequent term would be $ \cos^4 x = \left(\frac{1}{\sqrt{2}}\right)^4 = \frac{1}{4} $. The series becomes:

$ 1 + \frac{1}{\sqrt{2}} + \left(\frac{1}{\sqrt{2}}\right)^2 + \left(\frac{1}{\sqrt{2}}\right)^3 + \left(\frac{1}{\sqrt{2}}\right)^4 + \dots $

Identifying the Geometric Series

The series evaluated at $x = \frac{\pi}{4}$ takes the form $ 1 + r + r^2 + r^3 + r^4 + \dots $, where $ r = \frac{1}{\sqrt{2}} $. This is an infinite geometric series.

• The first term is $ a = 1 $.
• The common ratio is $ r = \frac{1}{\sqrt{2}} $.

Calculating the Sum

Since the absolute value of the common ratio $ |r| = |\frac{1}{\sqrt{2}}| < 1 $, the geometric series converges.

The sum $S$ of an infinite geometric series is given by the formula:

$ S = \frac{a}{1-r} $

Substituting the values $a=1$ and $r=\frac{1}{\sqrt{2}}$:

$ S = \frac{1}{1 - \frac{1}{\sqrt{2}}} $

To simplify, find a common denominator for the expression in the denominator:

$ S = \frac{1}{\frac{\sqrt{2}-1}{\sqrt{2}}} $

Now, invert and multiply:

$ S = \frac{\sqrt{2}}{\sqrt{2}-1} $

Thus, the value of the series at $x = \frac{\pi}{4}$ is $ \frac{\sqrt{2}}{\sqrt{2}-1} $. This corresponds to Option D.