Question

If $\sin (A+B)=1$ and $\cos (A-B)=1$, ${{0}^{\circ }}\le \left( A+B \right)\le {{90}^{\circ }}$ and $A\ge B$, then find A and B.

Answer 1

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: $\sin (A+B)=1$ and $\cos (A-B)=1$

Let us consider the first equation,

$\sin (A+B)=1$

We know that, $\sin {{90}^{\circ }}=1$, on comparing it with the above equation, we get,

$\sin \left( A+B \right)=\sin {{90}^{\circ }}$

Removing sine function from both the sides, we get,

$A+B={{90}^{\circ }}..........................(i)$

Now, let us consider the second equation,

$\cos (A-B)=1$

We know that, $\cos {{0}^{\circ }}=1$, on comparing it with the above equation, we get,

$\cos \left( A-B \right)=\cos {{0}^{\circ }}$

Removing cosine function from both the sides, we get,

$A-B={{0}^{\circ }}..........................(ii)$

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={{0}^{\circ }} \\

 & \Rightarrow B={{45}^{\circ }} \\

\end{align}$

Hence, angle A and B are both 45 degrees.


Note: One may note that, there are many angles for which the value of sine and cosine of the angle is 1 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.