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Q4(i):
State whether the following are true or false. Justify your answer. (i) $\sin (A + B) = \sin A + \sin B$.
Solution :
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 $\sin 30^\circ + \sin 60^\circ = \frac{1+\sqrt{3}}{2}$.
More Questions from Class 10 Mathematics Introduction to Trigonometry EXERCISE 8.2
- Q1(i): Evaluate the following : (i) $\sin 60^\circ \cos 30^\circ + \sin 30^\circ \cos 60^\circ$
- Q1(ii): Evaluate the following : (ii) $2 \tan^2 45^\circ + \cos^2 30^\circ – \sin^2 60^\circ$
- Q1(iii): Evaluate the following : (iii) $\frac{\cos 45^\circ}{\sec 30^\circ + \text{cosec } 30^\circ}$
- Q1(iv): Evaluate the following : (iv) $\frac{\sin 30^\circ + \tan 45^\circ – \text{cosec } 60^\circ}{\sec 30^\circ + \cos 60^\circ + \cot 45^\circ}$
- Q1(v): Evaluate the following : (v) $\frac{5 \cos^2 60^\circ + 4 \sec^2 30^\circ - \tan^2 45^\circ}{\sin^2 30^\circ + \cos^2 30^\circ}$
- Q2(i): Choose the correct option and justify your choice : (i) $\frac{2 \tan 30^\circ}{1 + \tan^2 30^\circ} =$
- Q2(ii): Choose the correct option and justify your choice : (ii) $\frac{1 - \tan^2 45^\circ}{1 + \tan^2 45^\circ} =$
- Q2(iii): Choose the correct option and justify your choice : (iii) $\sin 2A = 2 \sin A$ is true when $A =$
- Q2(iv): Choose the correct option and justify your choice : (iv) $\frac{2 \tan 30^\circ}{1 - \tan^2 30^\circ} =$
- Q3: If $\tan (A + B) = \sqrt{3}$ and $\tan (A – B) = \frac{1}{\sqrt{3}}$; $0^\circ < A + B \le 90^\circ$; $A > B$, find $A$ and $B$.
- Q4(ii): State whether the following are true or false. Justify your answer. (ii) The value of $\sin \theta$ increases as $\theta$ increases.
- Q4(iii): State whether the following are true or false. Justify your answer. (iii) The value of $\cos \theta$ increases as $\theta$ increases.
- Q4(iv): State whether the following are true or false. Justify your answer. (iv) $\sin \theta = \cos \theta$ for all values of $\theta$.
- Q4(v): State whether the following are true or false. Justify your answer. (v) $\cot A$ is not defined for $A = 0^\circ$.
CBSE Solutions for Class 10 Mathematics Introduction to Trigonometry
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