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Question: If we have a trigonometric expression \[2\cos (A+B)=2\sin (A-B)=1\]; find the values of \[A\] and \[...

If we have a trigonometric expression 2cos(A+B)=2sin(AB)=12\cos (A+B)=2\sin (A-B)=1; find the values of AA and BB.
A. A=40,B=20A=40,B=20
B. A=10,B=20A=10,B=20
C. A=40,B=25A=40,B=25
D. A=45,B=15A=45,B=15

Explanation

Solution

The given question has an expression given to us. We will have to solve the given expression and then find the values of AA and BB. We will be separating the equalities in the given expression and form two equations which are as follows, A+B=60A+B={{60}^{\circ }} and AB=30A-B={{30}^{\circ }}. We will then solve both the equations and find the value of AA and BB. Hence, we will have the required values of AA and BB.

Complete step by step solution:
According to the given question, we are given an expression using which we have to find the values of AA and BB.
The given expression we have is,
2cos(A+B)=2sin(AB)=12\cos (A+B)=2\sin (A-B)=1---(1)
We will first separate the equalities given to us, that is, we have,
2cos(A+B)=12\cos (A+B)=1 and 2sin(AB)=12\sin (A-B)=1
We will now take the above expressions individually and make equations using them. So, we have,
2cos(A+B)=12\cos (A+B)=1
cos(A+B)=12\Rightarrow \cos (A+B)=\dfrac{1}{2}
cos(A+B)=cos60\Rightarrow \cos (A+B)=\cos {{60}^{\circ }}
We can write the above expression as,
A+B=60\Rightarrow A+B={{60}^{\circ }}----(2)
Next, we have,
2sin(AB)=12\sin (A-B)=1
sin(AB)=12\Rightarrow \sin (A-B)=\dfrac{1}{2}
sin(AB)=sin30\Rightarrow \sin (A-B)=\sin {{30}^{\circ }}
We can write it as,
AB=30\Rightarrow A-B={{30}^{\circ }}-----(3)
We will now solve the equations (3) and (4).
From the equation (3), we can write, B=60AB={{60}^{\circ }}-A----(5)
substituting this in the equation (4), we get,
A(60A)=30\Rightarrow A-\left( {{60}^{\circ }}-A \right)={{30}^{\circ }}
Opening up the brackets, we have,
A60+A=30\Rightarrow A-{{60}^{\circ }}+A={{30}^{\circ }}
A+A=30+60\Rightarrow A+A={{30}^{\circ }}+{{60}^{\circ }}
Adding up the similar terms, we get,
2A=90\Rightarrow 2A={{90}^{\circ }}
So, we get the value of A as,
A=45\Rightarrow A={{45}^{\circ }}
We will now substitute the value of A in equation (5), we get,
B=6045\Rightarrow B={{60}^{\circ }}-{{45}^{\circ }}
So, we get the value of B as,
B=15\Rightarrow B={{15}^{\circ }}
Therefore, the value of A=45A={{45}^{\circ }} and B=15B={{15}^{\circ }}.
So, the option D. A=45,B=15A=45,B=15 is correct.

Note: The expression given to us consists of two different expressions, to convert them into further two equations, the trigonometric function with their values at a particular angle should be correctly known, else the equations formed will be wrong. Also, while substituting back the value of ‘A’ in the equation (5), the value should be correctly computed.