If $ A=30^{\circ} $ and $ B=60^{\circ} $, verify that$ \cos (A+B)=\cos A \cos B-\sin A \sin B $

Given:

\( A=30^{\circ} \) and \( B=60^{\circ} \)

To do:

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

Solution:  

We know that,

$\sin 30^{\circ}=\frac{1}{2}$

$\cos 30^{\circ}=\frac{\sqrt3}{2}$

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

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

Let us consider LHS,

$\cos (A+B)=\cos (30^{\circ}+60^{\circ})$

$=\cos 90^{\circ}$      

$=0$      (Since $\cos 90^{\circ}=0$)

Let us consider RHS,

$\cos A \cos B-\sin A \sin B=\cos 30^{\circ} \cos 60^{\circ}-\sin 30^{\circ} \sin 60^{\circ}$

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

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

$=0$

LHS $=$ RHS

Hence proved.   

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Updated on: 10-Oct-2022

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