Question

Prove that cos(3pi/4-A)×cos(3pi/4-B) - sin(3pi/4-A)×sin(3pi/4-B)=-sin(A+B)

Answer 1

To prove:

$cos(\dfrac{3\pi}{4}-A)\times cos(\dfrac{3\pi}{4}-B) - sin(\dfrac{3\pi}{4}-A)\times sin(\dfrac{3\pi}{4}-B)=-sin(A+B)$

LHS (Left Hand Side):

$cos(\dfrac{3\pi}{4}-A)\times cos(\dfrac{3\pi}{4}-B) - sin(\dfrac{3\pi}{4}-A)\times sin(\dfrac{3\pi}{4}-B)$

Formula:

$cosCcosD - SinCSinD = cos(C+D) ....... (1)$

$Let\ (\dfrac{3\pi}{4}-A) = C\\and\ (\dfrac{3\pi}{4}-B) = D$

The given equation becomes comparable to the formula given in equation(1).

So, the LHS of given equation becomes:

$cos(\dfrac{3\pi}{4}-A+\dfrac{3\pi}{4}-B)\\\Rightarrow cos(\dfrac{3\pi}{4}+\dfrac{3\pi}{4}-A-B)\\\Rightarrow cos((\dfrac{3\pi+3\pi}{4})-(A+B))\\\Rightarrow cos(\dfrac{6\pi}{4}-(A+B))\\\Rightarrow cos(\dfrac{3\pi}{2}-(A+B)) ...... (2)$

Formula:

$cos(\dfrac{3\pi}{2} - \theta) = -sin\theta$  (Because Cos changes to Sin and it is 2nd quadrant so negative value)

Comparing above formula with equation (2):

$\theta = (A+B)$

Equation (2) becomes:

$-sin(A+B) = RHS$ (Right Hand Side)

So, $cos(\dfrac{3\pi}{4}-A)\times cos(\dfrac{3\pi}{4}-B) - sin(\dfrac{3\pi}{4}-A)\times sin(\dfrac{3\pi}{4}-B)=-sin(A+B)$

Hence proved.