Express 4 sin(x) – 8 cos(x) in the form R sin(x-a), where R and a are constants, R >0 and 0< a< π/2

4 sin(x) – 8 cos(x)= Rsin(x-a) here use double angle formula

4 sin(x) – 8 cos(x)= Rsin(x)cos(a)-Rcos(x)sin(a) Rearrange so in same format as LHS

4 sin(x) – 8 cos(x)= Rcos(a)sin(x)-Rsin(a)cos(x) Equate terms of sin(x) and cos(x)

4=Rcos(a) 8= Rsin(a) Equal to R and so each other gives

4/cos(a)=8/sin(a) Rearranging to give tan identity (sin(a)/cos(a)=tans(a))

sin(a)/ cos(a)= 8/4

tan (a)=2 Remembering to give answer in radians as this is what the Q asks for

a=1.11 Sub this value into an original equation for R

R= 4/cos(a)= 4/cos(1.1)= 8.94

4 sin(x) – 8 cos(x)= 8.94sin(x-1.11)

Can check by plugging any value in e.g.π gives 8 and 8.008