To prove the given equation [tex]\(\sqrt{1+\sin \theta} = \cos \frac{\theta}{2} + \sin \frac{\theta}{2}\)[/tex], we will start by comparing both sides of the equation.
### Step-by-Step Solution:
#### 1. Expression on LHS
The left-hand side (LHS) of the equation is:
[tex]\[ \sqrt{1 + \sin \theta} \][/tex]
#### 2. Expression on RHS
The right-hand side (RHS) of the equation is:
[tex]\[ \cos \frac{\theta}{2} + \sin \frac{\theta}{2} \][/tex]
#### 3. Simplifying and Comparing
We need to check if the LHS expression is equivalent to the RHS expression.
#### 4. Simplified Form Check
To verify this:
Simplifying the expression, we find:
[tex]\[ \cos \frac{\theta}{2} + \sin \frac{\theta}{2} = \sqrt{1 + \sin \theta} - \sqrt{2} \sin \left( \frac{\theta}{2} + \frac{\pi}{4} \right) \][/tex]
Since [tex]\(\cos \frac{\theta}{2} + \sin \frac{\theta}{2}\)[/tex] simplifies to an expression involving a trigonometric function, we observe:
[tex]\[ \sqrt{1 + \sin \theta} \not= \cos \frac{\theta}{2} + \sin \frac{\theta}{2} \][/tex]
#### 5. Conclusion:
Upon closer examination, we see that the simplified difference between the two expressions [tex]\( \sqrt{1 + \sin \theta} - \sqrt{2} \sin \left( \frac{\theta}{2} + \frac{\pi}{4} \right) \)[/tex] is not equal to zero. Therefore, the LHS and RHS are not equal.
Thus, we conclude that:
[tex]\[ \sqrt{1 + \sin \theta} \neq \cos \frac{\theta}{2} + \sin \frac{\theta}{2} \][/tex]
So, the given equation:
[tex]\[ \sqrt{1+\sin \theta}=\cos \frac{\theta}{2}+\sin \frac{\theta}{2} \][/tex]
is not true.
#### Summary:
By simplifying and closely comparing both sides of the given equation, we found that the equation does not hold true.
