Q4(i):

State whether the following are true or false. Justify your answer. (i) $\sin (A + B) = \sin A + \sin B$.

Given: The trigonometric statement $\sin(A + B) = \sin A + \sin B$.

To Find: Determine whether the given statement is True or False and provide a justification.

Visual Representation:

Step 1: Understanding the nature of the trigonometric function
The expression $\sin(A + B)$ represents the sine of the sum of two angles $A$ and $B$. In trigonometry, the sine function is a non-linear operator. The distributive property $f(x+y) = f(x) + f(y)$ does not apply to trigonometric functions.

Step 2: Testing the statement with specific values
To verify if the statement is true for all values of $A$ and $B$, we can choose standard angles from the trigonometric table, such as $A = 30^\circ$ and $B = 60^\circ$.

Step 3: Calculating the Left-Hand Side (LHS)
LHS = $\sin(A + B)$
Substitute $A = 30^\circ$ and $B = 60^\circ$:
LHS = $\sin(30^\circ + 60^\circ)$
LHS = $\sin(90^\circ)$
[Since $\sin(90^\circ) = 1$ from the standard trigonometric ratio table]
LHS = $1$

Step 4: Calculating the Right-Hand Side (RHS)
RHS = $\sin A + \sin B$
Substitute $A = 30^\circ$ and $B = 60^\circ$:
RHS = $\sin(30^\circ) + \sin(60^\circ)$
[Using the values $\sin(30^\circ) = \frac{1}{2}$ and $\sin(60^\circ) = \frac{\sqrt{3}}{2}$]
RHS = $\frac{1}{2} + \frac{\sqrt{3}}{2}$
RHS = $\frac{1 + \sqrt{3}}{2}$

Step 5: Comparing LHS and RHS
We observe that:
$1 \neq \frac{1 + \sqrt{3}}{2}$
[Since $\sqrt{3} \approx 1.732$, then $\frac{1 + 1.732}{2} = \frac{2.732}{2} = 1.366$]
Since $1 \neq 1.366$, the LHS is not equal to the RHS.

Conclusion:
Because the equality does not hold for the chosen values of $A$ and $B$, the general statement $\sin(A + B) = \sin A + \sin B$ is mathematically incorrect.

Final Answer: False. The statement is incorrect because the sine function does not distribute over addition. As demonstrated with $A=30^\circ$ and $B=60^\circ$, $\sin(30^\circ+60^\circ) = 1$, whereas $\sin 30^\circ + \sin 60^\circ = \frac{1+\sqrt{3}}{2}$.