Question

prove that sin[45 +theta] -cos[45 -theta] =0

Answer 1

Answer:

Sin ( 45 + A ) = Sin 45 . Cos A + Cos 45 . Sin A   => Equation 1

Cos ( 45 - A ) = Cos 45 . Cos A + Sin 45 . Sin A   => Equation 2

According to the question,

To prove, Equation 1 - Equation 2 = 0

=> Sin 45 . Cos A + Cos 45. Sin A - Cos 45 . Cos A - Sin 45 . Sin A = 0

Evaluating, Cos 45 and Sin 45 as 1 / √2 we get,

=> 1 / √2 * Cos A + 1 / √2 * Sin A - 1 / √2 . Cos A - 1 / √2 . Sin A = 0

Taking 1 / √2 common we get,

=> 1 / √2 [ Cos A + Sin A - Cos A - Sin A ] = 0

=> [ 1 / √2 ] [ 0 ] = 0

=> 0 = 0

Hence LHS = RHS

Hence proved !

Answer 2

Given,

$\sin(45\textdegree+\theta)-\cos(45\textdegree-\theta)\\\;\\=\cos(90\textdegree-(45\textdegree+\theta))-\cos(45\textdegree-\theta)\\\;\\=\cos(90\textdegree-45\textdegree-\theta)-\cos(45\textdegree-\theta)\\\;\\=\cos(45\textdegree-\theta)-\cos(45\textdegree-\theta)\\\;\\=0\\\;\\\;\\Note:\;\;\sin\theta=\cos(90\textdegree-\theta)$