show that (sin A + cos A) (Sin B + Cos B)= Sin (A +B) + Cos (A-B)

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- Jan 28th 2009, 03:52 PMdanetrig question
show that (sin A + cos A) (Sin B + Cos B)= Sin (A +B) + Cos (A-B)

- Jan 28th 2009, 03:56 PMMush
$\displaystyle RHS = \sin(A+B) + \cos(A-B) $

Expand both of these:

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

Now factor out sin(A) and cos(A) from the relevant expressions:

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

Now realise that the inside of the brackets of both sides are identical. Hence the expression in the brackets is a factor of both terms. Take it our as a factor:

$\displaystyle =(\cos(B)+\sin(B))(\sin(A)+\cos(A)) = LHS $