Question

If we are given the expression as $\sin \left( \alpha -\beta \right)=\dfrac{1}{2}$ and $\cos \left( \alpha +\beta \right)=\dfrac{1}{2}$ , where $\alpha$ and $\beta$ are positive acute angles, then
(A) $\alpha ={{45}^{\circ }},\beta ={{15}^{\circ }}$
(B) $\alpha ={{15}^{\circ }},\beta ={{45}^{\circ }}$
(C) $\alpha ={{60}^{\circ }},\beta ={{15}^{\circ }}$
(D) None of these

Answer 1

In this question we have been asked to find the value of $\alpha$ and $\beta$ when $\sin \left( \alpha -\beta \right)=\dfrac{1}{2}$ and $\cos \left( \alpha +\beta \right)=\dfrac{1}{2}$ , they are positive acute angles. For doing that we will use the basic trigonometric values given as $\sin {{30}^{\circ }}=\dfrac{1}{2}=\cos {{60}^{\circ }}$ which we have learnt before.

Complete step-by-step solution:
Now considering from the question we have been asked to find the value of $\alpha$ and $\beta$ when $\sin \left( \alpha -\beta \right)=\dfrac{1}{2}$ and $\cos \left( \alpha +\beta \right)=\dfrac{1}{2}$ , they are positive acute angles.
For doing that we will use the basic trigonometric values given as $\sin {{30}^{\circ }}=\dfrac{1}{2}=\cos {{60}^{\circ }}$ which we have learnt before.
By using these values we can say that $\alpha -\beta ={{30}^{\circ }}$ and $\alpha +\beta ={{60}^{\circ }}$.
Now we will add both the equations to simplify them further. After doing that we will have $2\alpha ={{90}^{\circ }}$ .
Now we can conclude that the value of $\alpha$ is ${{45}^{\circ }}$ .
Now we will substitute this value in any one equation and simplify it for finding the value of the other variable.
By doing that we will have $\beta ={{15}^{\circ }}$ .
Therefore we can conclude that when $\sin \left( \alpha -\beta \right)=\dfrac{1}{2}$ and $\cos \left( \alpha +\beta \right)=\dfrac{1}{2}$ , then the values of $\alpha$ and $\beta$ are ${{45}^{\circ }}$ and ${{15}^{\circ }}$ respectively which are positive acute angles.
Hence we can mark the option “A” as correct.

Note: Now considering from the question we have been asked to find the value of $\alpha$ and $\beta$ when $\sin \left( \alpha -\beta \right)=\dfrac{1}{2}$ and $\cos \left( \alpha +\beta \right)=\dfrac{1}{2}$ , they are positive acute angles.
For doing that we will use the basic trigonometric values given as $\sin {{30}^{\circ }}=\dfrac{1}{2}=\cos {{60}^{\circ }}$ which we have learnt before.
By using these values we can say that $\alpha -\beta ={{30}^{\circ }}$ and $\alpha +\beta ={{60}^{\circ }}$.
Now we will add both the equations to simplify them further. After doing that we will have $2\alpha ={{90}^{\circ }}$ .
Now we can conclude that the value of $\alpha$ is ${{45}^{\circ }}$ .
Now we will substitute this value in any one equation and simplify it for finding the value of the other variable.
By doing that we will have $\beta ={{15}^{\circ }}$ .
Therefore we can conclude that when $\sin \left( \alpha -\beta \right)=\dfrac{1}{2}$ and $\cos \left( \alpha +\beta \right)=\dfrac{1}{2}$ , then the values of $\alpha$ and $\beta$ are ${{45}^{\circ }}$ and ${{15}^{\circ }}$ respectively which are positive acute angles.
Hence we can mark the option “A” as correct.