Question

Prove that COS (A-B) = COS A COS B +SIN A SIN B

Answer 1

Three steps solutions

Step-by-step explanation:

Prove that,

$cos(A - B) = cos A \times cos B + sin A\times sin B$

Taking L.H.S ,

$cos(A -B) = cos [A + (-B)]$

$cos(A -B) = cos A\times cos(-B) - sin A\times sin(-B)$

∵  $cos(-B) = cos B$ , $sin(-B) = -sin B$

∴  $cos(A - B) = cos A \times cos B - sin A\times (-sin B)$

Or,

$cos(A - B) = cos A\times cos B + sin A\times sin B$

Hence proved,

L.H.S. = R.H.S.

Answer 2

Hence proved

$\cos(A-B)=\cos A\cos B+\sin A\sin B$

Step-by-step explanation:

To prove: $\cos(A-B)=\cos A\cos B+\sin A\sin B$

As we have a formula,

$\cos(A+B)=\cos A\cos B-\sin A\sin B$

Replace B as -B

$\cos(A+(-B))=\cos A\cos (-B)-\sin A\sin (-B)$

$\cos(A-B)=\cos A\cos B-\sin A\sin (-B)$     $\because \cos(-x)=\cos x$

$\cos(A-B)=\cos A\cos B-\sin A(-\sin B)$     $\because \sin(-x)=-\sin x$

$\cos(A-B)=\cos A\cos B+\sin A\sin B$     $\because -(-)=+$

Hence proved

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