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Prove that $\cos (45^{\large\circ}-A).\cos(45^{\large\circ}-B)-\sin(45^{\large\circ}-A)\sin(45^{\large\circ}-B)=\sin(A+B)$

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Toolbox:
  • $\cos(A+B)=\cos A.\cos B-\sin A.\sin B$
  • $\cos(90-\theta)=\sin \theta$
LHS
$\cos (45^{\large\circ}-A).\cos(45^{\large\circ}-B)-\sin(45^{\large\circ}-A)\sin(45^{\large\circ}-B)$
$\Rightarrow \cos(45^{\large\circ}-A)+ (45^{\large\circ}-B)$
$\cos(A+B)=\cos A.\cos B-\sin A.\sin B$
$\Rightarrow \cos(90^{\large\circ}-(A+B))$
$\cos(90-\theta)=\sin \theta$
$\Rightarrow \sin(A+B)$=RHS
Hence proved
answered May 21, 2014 by sreemathi.v
 

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