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

Given:

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

To do:

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

Solution:  

We know that,

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

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

Let us consider LHS,

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

$=\sin 0^{\circ}$      

$=0$      (Since $\sin 0^{\circ}=0$)

Let us consider RHS,

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

$=\left(\frac{\sqrt3}{2}\right) \times \left(\frac{1}{2}\right) -\left(\frac{1}{2}\right) \times \left(\frac{\sqrt{3}}{2}\right)$

$=\frac{\sqrt3}{4} -\frac{\sqrt3}{4}$

$=0$

LHS $=$ RHS

Hence proved.  

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Updated on: 10-Oct-2022

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