
If $\sin (A-B)=\dfrac{1}{2}$ and $\cos (A+B)=\dfrac{1}{2}$, ${{0}^{\circ }}\le \left( A+B \right)\le {{90}^{\circ }}$ and $A\ge B$, then find A and B.
Answer
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Hint: Find the value of angle for which sine and cosine of the required angle is 1. Now, write them in the form of $\sin a=\sin b$ and $\cos a=\cos b$. Remove the trigonometric functions from both the sides and equate the angles. Form two linear equations in A and B. solve these two equations to determine the value of A and B.
Complete step-by-step answer:
We have been provided with two trigonometric equations given as: $\sin (A-B)=\dfrac{1}{2}$ and $\cos (A+B)=\dfrac{1}{2}$.
Let us consider the first equation,
$\sin (A-B)=\dfrac{1}{2}$
We know that, $\sin {{30}^{\circ }}=\dfrac{1}{2}$, on comparing it with the above equation, we get,
$\sin \left( A-B \right)=\sin {{30}^{\circ }}$
Removing sine function from both the sides, we get,
$A-B={{30}^{\circ }}..........................(i)$
Now, let us consider the second equation,
$\cos (A+B)=\dfrac{1}{2}$
We know that, $\cos {{60}^{\circ }}=\dfrac{1}{2}$, on comparing it with the above equation, we get,
$\cos \left( A+B \right)=\cos {{60}^{\circ }}$
Removing cosine function from both the sides, we get,
$A+B={{60}^{\circ }}..........................(ii)$
Now, we have two linear equations in A and B. Let us solve these equations.
Adding equation (i) and (ii), we get,
$\begin{align}
& 2A={{90}^{\circ }} \\
& \Rightarrow A={{45}^{\circ }} \\
\end{align}$
Substituting, $A={{45}^{\circ }}$ in equation (ii), we get,
\[\begin{align}
& {{45}^{\circ }}+B={{60}^{\circ }} \\
& \Rightarrow B={{15}^{\circ }} \\
\end{align}\]
Hence, angle A is 45 degrees and angle B is 15 degrees.
Note: One may note that, there are many angles for which the value of sine and cosine of the angle is $\dfrac{1}{2}$ because these trigonometric functions are repeating in nature. One can see that we have been provided with the condition: ${{0}^{\circ }}\le \left( A+B \right)\le {{90}^{\circ }}$. So, we must select the angles from this range only. These angles, A and B, lie in the first quadrant only.
Complete step-by-step answer:
We have been provided with two trigonometric equations given as: $\sin (A-B)=\dfrac{1}{2}$ and $\cos (A+B)=\dfrac{1}{2}$.
Let us consider the first equation,
$\sin (A-B)=\dfrac{1}{2}$
We know that, $\sin {{30}^{\circ }}=\dfrac{1}{2}$, on comparing it with the above equation, we get,
$\sin \left( A-B \right)=\sin {{30}^{\circ }}$
Removing sine function from both the sides, we get,
$A-B={{30}^{\circ }}..........................(i)$
Now, let us consider the second equation,
$\cos (A+B)=\dfrac{1}{2}$
We know that, $\cos {{60}^{\circ }}=\dfrac{1}{2}$, on comparing it with the above equation, we get,
$\cos \left( A+B \right)=\cos {{60}^{\circ }}$
Removing cosine function from both the sides, we get,
$A+B={{60}^{\circ }}..........................(ii)$
Now, we have two linear equations in A and B. Let us solve these equations.
Adding equation (i) and (ii), we get,
$\begin{align}
& 2A={{90}^{\circ }} \\
& \Rightarrow A={{45}^{\circ }} \\
\end{align}$
Substituting, $A={{45}^{\circ }}$ in equation (ii), we get,
\[\begin{align}
& {{45}^{\circ }}+B={{60}^{\circ }} \\
& \Rightarrow B={{15}^{\circ }} \\
\end{align}\]
Hence, angle A is 45 degrees and angle B is 15 degrees.
Note: One may note that, there are many angles for which the value of sine and cosine of the angle is $\dfrac{1}{2}$ because these trigonometric functions are repeating in nature. One can see that we have been provided with the condition: ${{0}^{\circ }}\le \left( A+B \right)\le {{90}^{\circ }}$. So, we must select the angles from this range only. These angles, A and B, lie in the first quadrant only.
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