Question

cos(pie/4_x)cos(pie/4-y) -sin(pie/4-x) sin(pie/4-y) =

Answer 1

cos(π/4-x+π/4-x)=cos(2π/4)
cos ( π/2)=0

Answer 2

$cos(\dfrac{\pi}{4}-x).cos(\dfrac{\pi}{4}-y)-sin(\dfrac{\pi}{4}-x).sin(\dfrac{\pi}{4}-y)=sin(x+y)$

Explanation:

The given expression is as follows :

$cos(\dfrac{\pi}{4}-x).cos(\dfrac{\pi}{4}-y)-sin(\dfrac{\pi}{4}-x).sin(\dfrac{\pi}{4}-y)$

Using trigonometric identity :

$cosAcosB-sinAsinB=cos(A+B)$

Here, $A=\dfrac{\pi}{4}-x$

and

$B=\dfrac{\pi}{4}-y$

On using above identity, we get :

$cos(\dfrac{\pi}{4}-x).cos(\dfrac{\pi}{4}-y)-sin(\dfrac{\pi}{4}-x).sin(\dfrac{\pi}{4}-y)=cos(\dfrac{\pi}{2}-(x+y))$

$cos(\dfrac{\pi}{4}-x).cos(\dfrac{\pi}{4}-y)-sin(\dfrac{\pi}{4}-x).sin(\dfrac{\pi}{4}-y)=sin(x+y)$

Hence, this is the required solution.

Learn more :

Topic : Trigonometric identity

https://brainly.in/question/11930969