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Home  >>  CBSE XI  >>  Math  >>  Trigonometric Functions
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Prove that $\cos^2A+\cos ^2B-2\cos A\cos B.\cos(A+B)=\sin^2(A+B)$

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Toolbox:
  • $\cos^2A-\cos^2B=\cos(A+B)-\cos(A-B)$
  • $\cos(A-B)=\cos A.\cos B+\sin A.\sin B$
  • $\cos(A+B)=\cos A.\cos B-\sin A.\sin B$
LHS
$\cos^2A+\cos ^2B-2\cos A\cos B.\cos(A+B)$
$\cos^2A+(1-\sin ^2B)-2\cos A\cos B.\cos(A+B)$
$\cos^2B=1-\sin ^2B$
$1+\cos^2A-\sin ^2B-2\cos A\cos B\cos(A+B)$
$1+\cos(A+B)\cos(A-B)-2\cos A\cos B\cos(A+B)$
$\cos^2A-\cos^2B=\cos(A+B)-\cos(A-B)$
$1+\cos(A+B)[\cos(A-B)-2\cos A\cos B]$
$1+\cos(A+B)[\cos A\cos B+\sin A.\sin B)-2\cos A\cos B]$
$\cos(A-B)=\cos A.\cos B+\sin A.\sin B$
$1+\cos(A+B)(\sin A\sin B-\cos A\cos B)$
$1-\cos(A+B)(\cos A\cos B-\sin A\sin B)$
$\cos(A+B)=\cos A.\cos B-\sin A.\sin B$
$1-\cos^2(A+B)$
$\sin^2(A+B)$=RHS
Hence proved
answered May 21, 2014 by sreemathi.v
 

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