Question

Question #d76e9

Answer 1

`0`

Explanation:

`sin2x=1`, `2x=pi/2` or `x=pi/4`

Hence

`u=cos(8*pi/4)+2cos(6*pi/4)+cos(4*pi/4)`

`=cos(2pi)+2cos((3pi)/2)+cos(pi)`

`=1+2*0-1`

`=0`

1) I found `x` for `sin2x=1` condition.

2) I substituted to `x=pi/4`.

Answer 2

`0`.

Explanation:

`sin(2x)=1`.

In general, equations of the form `sin(theta)=K` and solved by `theta = arcsin(K)+2npi` and `pi-theta=arcsin(K)+2npi`, `n in ZZ`, because of the periodicity of the sine function#.

Note that `arcsin(1)=pi/2`. This means that if `sin(2x)=1`,

`2x=pi/2+2npi`,
`x=pi/4+npi`.

`pi-2x=pi/2+2npi`,
`2x=pi/2-2npi`,
`x=pi/4-npi`.

As `n in ZZ` these solution sets are identical.
So if `sin(2x)=1` then `x=pi/4+npi`, `n in ZZ`.

Then

`cos(8(pi/4+npi)+2cos(6(pi/4+npi))+cos(4(pi/4+npi))`,
`cos(2pi+4(2npi))+2cos((3pi)/2+3(2npi))+cos(pi+2(2npi))`.

Multiplies of the periodicity can be removed and the values substituted to give,

`1+2*0-1=0`.

In this case, it was safe to assume `sin(2x)=1 => x=pi/4`. This will not always be the case! So it is good to solve more generally.