Question

Given $ A={{60}^{\circ }} $ and $ B={{30}^{\circ }} $ , verify that: $ \cos \left( A+B \right)=\cos A.\cos B-\sin A.\sin B $ ?

Answer 1

Hint: We start solving the problem by considering the L.H.S (Left Hand Side) of the given equation. We then substitute the given values of A and B in it and perform the necessary calculations to get its value. We then consider the R.H.S (Right Hand Side) of the given equation and substitute the given values of A and B in it. We then make the necessary calculations to get its value. We then compare both values of L.H.S (Left Hand Side) and R.H.S (Right Hand Side) to complete the proof.

Complete step by step answer:
According to the problem, we need to verify $ \cos \left( A+B \right)=\cos A.\cos B-\sin A.\sin B $ if it is given $ A={{60}^{\circ }} $ and $ B={{30}^{\circ }} $ .
Let us first find the value of L.H.S (Left Hand Side).
So, we have $ \cos \left( A+B \right)=\cos \left( {{60}^{\circ }}+{{30}^{\circ }} \right) $ .
 $ \Rightarrow \cos \left( A+B \right)=\cos \left( {{90}^{\circ }} \right) $ .
 $ \Rightarrow \cos \left( A+B \right)=0 $ ---(1).
Now, let us find the value of R.H.S (Right-Hand side).
So, we have $ \cos A.\cos B-\sin A.\sin B $ .
 $ \Rightarrow \cos A.\cos B-\sin A.\sin B=\cos \left( {{60}^{\circ }} \right).\cos \left( {{30}^{\circ }} \right)-\sin \left( {{60}^{\circ }} \right).\sin \left( {{30}^{\circ }} \right) $ ---(2).
We know that $ \cos \left( {{60}^{\circ }} \right)=\dfrac{1}{2} $ , $ \cos \left( {{30}^{\circ }} \right)=\dfrac{\sqrt{3}}{2} $ , $ \sin \left( {{60}^{\circ }} \right)=\dfrac{\sqrt{3}}{2} $ and $ \sin \left( {{30}^{\circ }} \right)=\dfrac{1}{2} $ . Let us substitute these results in equation (2).
 $ \Rightarrow \cos A.\cos B-\sin A.\sin B=\left( \dfrac{1}{2}\times \dfrac{\sqrt{3}}{2} \right)-\left( \dfrac{\sqrt{3}}{2}\times \dfrac{1}{2} \right) $ .
 $ \Rightarrow \cos A.\cos B-\sin A.\sin B=\left( \dfrac{\sqrt{3}}{4} \right)-\left( \dfrac{\sqrt{3}}{4} \right) $ .
 $ \Rightarrow \cos A.\cos B-\sin A.\sin B=0 $ ---(3).
From equations (1) and (3), we can see that the values of $ \cos \left( A+B \right) $ and $ \cos A.\cos B-\sin A.\sin B $ are equal. This means that we have proved L.H.S (Left Hand Side) = R.H.S (Right Hand Side).
 $ \therefore $ We have verified the given result $ \cos \left( A+B \right)=\cos A.\cos B-\sin A.\sin B $ by taking the values $ A={{60}^{\circ }} $ and $ B={{30}^{\circ }} $ .

Note:
 We can also solve the R.H.S (Right Hand Side) of the given equation as shown below:
So, we have $ \cos A.\cos B-\sin A.\sin B $ .
 $ \Rightarrow \cos A.\cos B-\sin A.\sin B=\cos \left( {{60}^{\circ }} \right).\cos \left( {{30}^{\circ }} \right)-\sin \left( {{60}^{\circ }} \right).\sin \left( {{30}^{\circ }} \right) $ .
\[\Rightarrow \cos A.\cos B-\sin A.\sin B=\cos \left( {{60}^{\circ }} \right).\cos \left( {{30}^{\circ }} \right)-\sin \left( {{90}^{\circ }}-{{30}^{\circ }} \right).\sin \left( {{90}^{\circ }}-{{60}^{\circ }} \right)\].
We know that $ \sin \left( {{90}^{\circ }}-\alpha \right)=\cos \alpha $ .
\[\Rightarrow \cos A.\cos B-\sin A.\sin B=\cos \left( {{60}^{\circ }} \right).\cos \left( {{30}^{\circ }} \right)-\cos \left( {{30}^{\circ }} \right).\cos \left( {{60}^{\circ }} \right)\].
\[\Rightarrow \cos A.\cos B-\sin A.\sin B=\cos \left( {{60}^{\circ }} \right).\cos \left( {{30}^{\circ }} \right)-\cos \left( {{60}^{\circ }} \right).\cos \left( {{30}^{\circ }} \right)\].
\[\Rightarrow \cos A.\cos B-\sin A.\sin B=0\].