Question

under root of 1+ sin theta by binomial expansion

Answer 1

1 + \frac{1}{2}\sin \theta - \frac{1}{8}\sin^2 \theta + \frac{1}{16}\sin^3 \theta - \dots

Answer 2

The question asks for the binomial expansion of $\sqrt{1 + \sin \theta}$.

The binomial theorem for a real exponent $n$ is given by: $(1+x)^n = 1 + nx + \frac{n(n-1)}{2!}x^2 + \frac{n(n-1)(n-2)}{3!}x^3 + \dots$ This expansion is valid for $|x| < 1$.

In this problem, we have $\sqrt{1 + \sin \theta}$, which can be written as $(1 + \sin \theta)^{1/2}$. Here, $x = \sin \theta$ and $n = \frac{1}{2}$. For the expansion to be valid, we require $|\sin \theta| < 1$, which means $\theta \ne (k + 1/2)\pi$ for any integer $k$.

Let's compute the first few terms of the expansion:

1. First term: $1$
2. Second term: $nx = \frac{1}{2}(\sin \theta)$
3. Third term: $\frac{n(n-1)}{2!}x^2 = \frac{\frac{1}{2}(\frac{1}{2}-1)}{2}\sin^2 \theta = \frac{\frac{1}{2}(-\frac{1}{2})}{2}\sin^2 \theta = \frac{-\frac{1}{4}}{2}\sin^2 \theta = -\frac{1}{8}\sin^2 \theta$
4. Fourth term: $\frac{n(n-1)(n-2)}{3!}x^3 = \frac{\frac{1}{2}(\frac{1}{2}-1)(\frac{1}{2}-2)}{6}\sin^3 \theta = \frac{\frac{1}{2}(-\frac{1}{2})(-\frac{3}{2})}{6}\sin^3 \theta = \frac{\frac{3}{8}}{6}\sin^3 \theta = \frac{3}{48}\sin^3 \theta = \frac{1}{16}\sin^3 \theta$

Combining these terms, the binomial expansion of $\sqrt{1 + \sin \theta}$ is:

The binomial expansion of $\sqrt{1 + \sin \theta}$ is: $\sqrt{1 + \sin \theta} = 1 + \frac{1}{2}\sin \theta - \frac{1}{8}\sin^2 \theta + \frac{1}{16}\sin^3 \theta - \dots$

Note on common simplification: In the context of competitive exams like JEE/NEET, expressions like $\sqrt{1 + \sin \theta}$ are frequently simplified using trigonometric identities to a closed form. This simplification is as follows: We use the identities $\sin^2 x + \cos^2 x = 1$ and $\sin 2x = 2 \sin x \cos x$. Let $x = \frac{\theta}{2}$. Then $1 = \sin^2 \frac{\theta}{2} + \cos^2 \frac{\theta}{2}$ and $\sin \theta = 2 \sin \frac{\theta}{2} \cos \frac{\theta}{2}$. Substituting these into the expression: $1 + \sin \theta = \left(\sin^2 \frac{\theta}{2} + \cos^2 \frac{\theta}{2}\right) + \left(2 \sin \frac{\theta}{2} \cos \frac{\theta}{2}\right)$ This is in the form of $(a^2 + b^2 + 2ab)$, which is the expansion of $(a+b)^2$. So, $1 + \sin \theta = \left(\sin \frac{\theta}{2} + \cos \frac{\theta}{2}\right)^2$. Taking the square root: $\sqrt{1 + \sin \theta} = \sqrt{\left(\sin \frac{\theta}{2} + \cos \frac{\theta}{2}\right)^2} = \left|\sin \frac{\theta}{2} + \cos \frac{\theta}{2}\right|$ While this closed-form expression is a common simplification, the question specifically asked for the "binomial expansion", which refers to the infinite series.