Question

Prove that `cos(y-pi)+sin(y+pi/3)=0`?

Answer 1

Sorry `cos(y-pi)+sin(y+pi/3)!=0`, but `cos(y-pi)+sin(y+pi/3)=-cosy+1/2siny+sqrt3/2cosy`

Explanation:

We use the formulas `cos(A-B)=cosAcosB+sinAsinB` and

`sin(A+B)=sinAcosB+cosAsinB`

Hence `cos(y-pi)+sin(y+pi/3)`

= `cosycospi+sinysinpi+sinycos(pi/3)+cosysin(pi/3)`

As `cospi=-1`, `sinpi=0`, `cos(pi/3)=1/2` and `sin(pi/3)=sqrt3/2`

`cosycospi+sinysinpi+sinycos(pi/3)+cosysin(pi/3)`

= `cosyxx(-1)+sinyxx0+sinyxx1/2+cosyxxsqrt3/2`

= `-cosy+1/2siny+sqrt3/2cosy`

Answer 2

Perhaps, the solution is required.
If so, #y = kpi+pi/12, k = 0. +-1, +-2, +-3...

Explanation:

`cos (y-pi)=cos(pi-y)=-cos y`
`sin (y+pi/3)= sin y cos (pi/3) + cos y sin (pi/3)=(1/2)(sin y + sqrt 3 cos y)`

So, the given equation becomes
`sin y = (2-sqrt3) cos y`
`tan y = 2-sqrt 3`
The principal value of `y = pi/12`.
General value of `y = kpi+pi/12, k = 0. +-1, +-2, +-3...`