If $ A=B=60^{\circ} $, verify that$ \cos (A-B)=\cos A \cos B+\sin A \sin B $

Given:

\( A=B=60^{\circ} \)

To do:

We have to verify that \( \cos (A-B)=\cos A \cos B+\sin A \sin B \).

Solution:  

We know that,

$\sin 60^{\circ}=\frac{\sqrt3}{2}$

$\cos 60^{\circ}=\frac{1}{2}$

Let us consider LHS,

$\cos (A-B)=\cos (60^{\circ}-60^{\circ})$

$=\cos 0^{\circ}$      

$=1$      (Since $\cos 0^{\circ}=1$

Let us consider RHS,

$\cos A \cos B+\sin A \sin B=\cos 60^{\circ} \cos 60^{\circ}+\sin 60^{\circ} \sin 60^{\circ}$

$=\left(\frac{1}{2}\right) \times \left(\frac{1}{2}\right) +\left(\frac{\sqrt{3}}{2}\right) \times \left(\frac{\sqrt{3}}{2}\right)$

$=\frac{1}{4} +\frac{3}{4}$

$=1$

LHS $=$ RHS

Hence proved. 

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Updated on: 10-Oct-2022

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