Question

$\sqrt{1 + \sin \theta}$ is equal to

• A. $\sin \theta + \cos \theta$
• B. $\sin \theta - \cos \theta$
• C. $\sin \frac{\theta}{2} + \cos \frac{\theta}{2}$
• D. $\sin \frac{\theta}{2} - \cos \frac{\theta}{2}$

Answer 1

Answer: (C)

$\sin \frac{\theta}{2} + \cos \frac{\theta}{2}$

Answer 2

To simplify the expression $\sqrt{1 + \sin \theta}$, we use fundamental trigonometric identities.

We know that:

1. The Pythagorean identity: $\sin^2 x + \cos^2 x = 1$.
2. The double angle identity for sine: $\sin 2x = 2 \sin x \cos x$.

Let's apply these identities by setting $x = \frac{\theta}{2}$.
Then, we can write $1$ as $\sin^2 \frac{\theta}{2} + \cos^2 \frac{\theta}{2}$, and we can write $\sin \theta$ as $2 \sin \frac{\theta}{2} \cos \frac{\theta}{2}$.

Substitute these into the expression $1 + \sin \theta$:

$1 + \sin \theta = \left(\sin^2 \frac{\theta}{2} + \cos^2 \frac{\theta}{2}\right) + \left(2 \sin \frac{\theta}{2} \cos \frac{\theta}{2}\right)$

This expression is in the form of $(a^2 + b^2 + 2ab)$, which is the expansion of $(a+b)^2$.
Here, $a = \sin \frac{\theta}{2}$ and $b = \cos \frac{\theta}{2}$.
So, $1 + \sin \theta = \left(\sin \frac{\theta}{2} + \cos \frac{\theta}{2}\right)^2$.

Now, we need to find the square root of this expression:

$\sqrt{1 + \sin \theta} = \sqrt{\left(\sin \frac{\theta}{2} + \cos \frac{\theta}{2}\right)^2}$

Therefore, $\sqrt{1 + \sin \theta} = \left|\sin \frac{\theta}{2} + \cos \frac{\theta}{2}\right|$.

Given the options, the most appropriate answer is the simplified form without the absolute value, assuming the principal root and a suitable domain for $\theta$.