Question

cos ( π/4-2) cos ( π/4-4)-sin(π/4-x) sin ( π/4-y)​

Answer 1

Step-by-step explanation:

given: cos(pi/4-2)cos(pi/4-4)-sin(pi/4-x)sin(pi/4-y)

we know that:

cos(A-B)= cosA.cosB+sinA.sinB

sin(A-B)= sinA.cosB-cosA.sinB

by substituting the values in the above two formulas.....

cos(pi/2-4)= cos pi/2.cos4+sin pi/2.sin4

we know that: cos pi/2=0, sin pi/2= 1

so, (0)cos4+(1)sin4= sin4

cos(pi/4-4)= cos pi/4.cos4+sin pi/4.sin4

we know that: cos pi/4=1/√2, sin pi/4= 1/√2

so, (1/√2)cos4+ (1/√2)sin4

= cos4+sin4/√2

sin(pi/4-x)= sin pi/4 cosX-cos pi/4 sinX

= cosX-sinX/√2

sin(pi/4-y)= sin pi/4 cosY- cos pi/4 sinY

= cosY-sinY/√2

so by multiplying all the values as Stated in the question, we have...

=(sin4)(cos4+sin4/√2)(cosX-sinX/√2)(cosY-sinY/√2)

hope it helps you friend....