# trig question

• January 28th 2009, 03:52 PM
dane
trig question
show that (sin A + cos A) (Sin B + Cos B)= Sin (A +B) + Cos (A-B)
• January 28th 2009, 03:56 PM
Mush
Quote:

Originally Posted by dane
show that (sin A + cos A) (Sin B + Cos B)= Sin (A +B) + Cos (A-B)

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

Expand both of these:

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

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

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

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:

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