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Home  >>  CBSE XI  >>  Math  >>  Trigonometric Functions
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Prove that $(\cos A+\cos B)^2+(\sin A+\sin B)^2=4\cos^2\large\frac{A-B}{2}$

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Toolbox:
  • $\sin^2A+\cos^2A=1$
  • $\cos^2\theta=\large\frac{1}{2}$$(1+\cos 2\theta)$
LHS
$(\cos A+\cos B)^2+(\sin A+\sin B)$
$\Rightarrow \cos^2A+\cos^2B+2\cos A\cos B+\sin^2A+\sin ^2B+2\sin A\sin B$
$\Rightarrow (\cos^2A+\sin ^2A)+(\cos^2B+\sin^2A)+2(\sin A.\sin B+\cos A.\cos B)$
$\Rightarrow 1+1+2(\cos(A-B))$
$\Rightarrow 2(1+\cos(A-B))$
$\Rightarrow 2\times 2\cos^2\large\frac{A-B}{2}=$$4\cos^2\large\frac{A-B}{2}$
Hence proved
answered May 22, 2014 by sreemathi.v
 

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