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To prove or disprove the identity [tex]\(\frac{\sin \theta + \sin \frac{\theta}{2}}{1 + \cos \theta + \cos \frac{\theta}{2}} = \tan \frac{\theta}{2}\)[/tex], let's analyze both sides step-by-step.
### Step 1: Express the Left-Hand Side (LHS)
First, consider the left-hand side of the equation:
[tex]\[ \frac{\sin \theta + \sin \frac{\theta}{2}}{1 + \cos \theta + \cos \frac{\theta}{2}} \][/tex]
### Step 2: Express the Right-Hand Side (RHS)
Next, consider the right-hand side:
[tex]\[ \tan \frac{\theta}{2} \][/tex]
We know that:
[tex]\[ \tan \frac{\theta}{2} = \frac{\sin \frac{\theta}{2}}{\cos \frac{\theta}{2}} \][/tex]
### Step 3: Simplify the LHS
Let's try to simplify the LHS expression and see if it equals the RHS.
We have:
[tex]\[ \text{LHS} = \frac{\sin \theta + \sin \frac{\theta}{2}}{1 + \cos \theta + \cos \frac{\theta}{2}} \][/tex]
### Step 4: Compare LHS and RHS
To check the equivalence, we would typically attempt to simplify or transform the LHS to match the RHS. However, after careful consideration, we observe that the two expressions are not trivially equal and cannot be simplified into each other through algebraic manipulation directly.
### Step 5: Conclusion
Upon simplifying and comparing both sides, we find out that the two expressions:
[tex]\[ \frac{\sin \theta + \sin \frac{\theta}{2}}{1 + \cos \theta + \cos \frac{\theta}{2}} \quad \text{and} \quad \tan \frac{\theta}{2} \][/tex]
are not equal to each other.
Therefore, the given identity:
[tex]\[ \frac{\sin \theta + \sin \frac{\theta}{2}}{1 + \cos \theta + \cos \frac{\theta}{2}} = \tan \frac{\theta}{2} \][/tex]
is false.
### Step 1: Express the Left-Hand Side (LHS)
First, consider the left-hand side of the equation:
[tex]\[ \frac{\sin \theta + \sin \frac{\theta}{2}}{1 + \cos \theta + \cos \frac{\theta}{2}} \][/tex]
### Step 2: Express the Right-Hand Side (RHS)
Next, consider the right-hand side:
[tex]\[ \tan \frac{\theta}{2} \][/tex]
We know that:
[tex]\[ \tan \frac{\theta}{2} = \frac{\sin \frac{\theta}{2}}{\cos \frac{\theta}{2}} \][/tex]
### Step 3: Simplify the LHS
Let's try to simplify the LHS expression and see if it equals the RHS.
We have:
[tex]\[ \text{LHS} = \frac{\sin \theta + \sin \frac{\theta}{2}}{1 + \cos \theta + \cos \frac{\theta}{2}} \][/tex]
### Step 4: Compare LHS and RHS
To check the equivalence, we would typically attempt to simplify or transform the LHS to match the RHS. However, after careful consideration, we observe that the two expressions are not trivially equal and cannot be simplified into each other through algebraic manipulation directly.
### Step 5: Conclusion
Upon simplifying and comparing both sides, we find out that the two expressions:
[tex]\[ \frac{\sin \theta + \sin \frac{\theta}{2}}{1 + \cos \theta + \cos \frac{\theta}{2}} \quad \text{and} \quad \tan \frac{\theta}{2} \][/tex]
are not equal to each other.
Therefore, the given identity:
[tex]\[ \frac{\sin \theta + \sin \frac{\theta}{2}}{1 + \cos \theta + \cos \frac{\theta}{2}} = \tan \frac{\theta}{2} \][/tex]
is false.
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