Which of the following values of A and B satisfies, sin(A + B) = sin A + sin B, where
A = 0° , B = 90°
Finding Values for the Trigonometric Equation sin(A + B) = sin A + sin B
The problem asks us to find the values of angles A and B that satisfy the given trigonometric equation:
$$ \sin(A + B) = \sin A + \sin B $$
We are given four pairs of values for A and B as options. We need to test each option to see which pair satisfies the equation.
Testing Option 1: A = 0°, B = 90°
Substitute A = 0° and B = 90° into the equation:
- Left side: $$ \sin(A + B) = \sin(0° + 90°) = \sin(90°) $$
- We know that $$ \sin(90°) = 1 $$
- Right side: $$ \sin A + \sin B = \sin(0°) + \sin(90°) $$
- We know that $$ \sin(0°) = 0 $$ and $$ \sin(90°) = 1 $$
- So, the right side is $$ 0 + 1 = 1 $$
Comparing both sides, we have:
$$ \text{Left side} = 1 $$
$$ \text{Right side} = 1 $$
Since Left side = Right side, the values A = 0° and B = 90° satisfy the equation $$ \sin(A + B) = \sin A + \sin B $$.
Testing Option 2: A = 45°, B = 45°
Substitute A = 45° and B = 45° into the equation:
- Left side: $$ \sin(A + B) = \sin(45° + 45°) = \sin(90°) $$
- We know that $$ \sin(90°) = 1 $$
- Right side: $$ \sin A + \sin B = \sin(45°) + \sin(45°) $$
- We know that $$ \sin(45°) = \frac{1}{\sqrt{2}} $$
- So, the right side is $$ \frac{1}{\sqrt{2}} + \frac{1}{\sqrt{2}} = \frac{2}{\sqrt{2}} = \sqrt{2} $$
Comparing both sides, we have:
$$ \text{Left side} = 1 $$
$$ \text{Right side} = \sqrt{2} $$
Since $$ 1 \neq \sqrt{2} $$, the values A = 45° and B = 45° do not satisfy the equation.
Testing Option 3: A = 30°, B = 30°
Substitute A = 30° and B = 30° into the equation:
- Left side: $$ \sin(A + B) = \sin(30° + 30°) = \sin(60°) $$
- We know that $$ \sin(60°) = \frac{\sqrt{3}}{2} $$
- Right side: $$ \sin A + \sin B = \sin(30°) + \sin(30°) $$
- We know that $$ \sin(30°) = \frac{1}{2} $$
- So, the right side is $$ \frac{1}{2} + \frac{1}{2} = 1 $$
Comparing both sides, we have:
$$ \text{Left side} = \frac{\sqrt{3}}{2} $$
$$ \text{Right side} = 1 $$
Since $$ \frac{\sqrt{3}}{2} \neq 1 $$, the values A = 30° and B = 30° do not satisfy the equation.
Testing Option 4: A = 60°, B = 30°
Substitute A = 60° and B = 30° into the equation:
- Left side: $$ \sin(A + B) = \sin(60° + 30°) = \sin(90°) $$
- We know that $$ \sin(90°) = 1 $$
- Right side: $$ \sin A + \sin B = \sin(60°) + \sin(30°) $$
- We know that $$ \sin(60°) = \frac{\sqrt{3}}{2} $$ and $$ \sin(30°) = \frac{1}{2} $$
- So, the right side is $$ \frac{\sqrt{3}}{2} + \frac{1}{2} = \frac{\sqrt{3} + 1}{2} $$
Comparing both sides, we have:
$$ \text{Left side} = 1 $$
$$ \text{Right side} = \frac{\sqrt{3} + 1}{2} $$
Since $$ 1 \neq \frac{\sqrt{3} + 1}{2} $$, the values A = 60° and B = 30° do not satisfy the equation.
Conclusion
After testing all the options, we found that only the values A = 0° and B = 90° satisfy the given trigonometric equation $$ \sin(A + B) = \sin A + \sin B $$.
Here is a summary of the results:
| A | B | sin(A + B) | sin A + sin B | Satisfies? |
|---|---|---|---|---|
| 0° | 90° | sin(90°) = 1 | sin(0°) + sin(90°) = 0 + 1 = 1 | Yes |
| 45° | 45° | sin(90°) = 1 | sin(45°) + sin(45°) = $ \frac{1}{\sqrt{2}} + \frac{1}{\sqrt{2}} = \sqrt{2} $ | No |
| 30° | 30° | sin(60°) = $ \frac{\sqrt{3}}{2} $ | sin(30°) + sin(30°) = $ \frac{1}{2} + \frac{1}{2} = 1 $ | No |
| 60° | 30° | sin(90°) = 1 | sin(60°) + sin(30°) = $ \frac{\sqrt{3}}{2} + \frac{1}{2} = \frac{\sqrt{3} + 1}{2} $ | No |
Therefore, the correct values for A and B are A = 0° and B = 90°.
Important Questions from Circular Measure of Angles
If tan 3 θ = cot(θ - 22°), where 3 θ is an angle, find the value of θ.
The value of 4cos\(\left( {\frac{\pi }{6}\, - \,\alpha } \right)\) sin\(\left( {\frac{\pi }{3}\, - \,\alpha } \right)\) is equal to:
1 + tan 15° cot 75° is equal to:
If tan θ = 1/√5, find the value of cosec2θ – sec2θ.
If cos A = \(\frac{63}{65}\), then find the value of tan A + cot A (up to two places of decimal).
