Question

1) If θ is a fixed real number with 0≤θ<2π Show that for all real x, cos(x)+cos(x+θ)=Acos(x+φ) where φ=θ/2 and A=2cos(θ/2) 2) Determine θ if A=1/4 and if A=-1?

I would really appreciate any help... Part should be related with compound angle formulae cos(α+β) and cos(α-β)

Answer 1

If `A=1/4`

for `0<=theta<=2pi` the solution is: `theta=2arccos(1/8)`

If `A=-1`
for `0<=theta<=2pi` the solution is: `theta=4pi/3`

Explanation:

Using prosthaphaeresis formula:
`cosalpha+cosbeta=2cos(alpha +beta)/2*cos(alpha -beta)/2`
could obtain:

`cos(x)+cos(x+θ)=2cos((x+x+theta)/2) *cos((x-x-theta)/2)`

`cos(x)+cos(x+θ)=2cos((x+x+theta)/2 *cos((x-x-theta)/2)`

`cos(x)+cos(x+θ)=2cos((2x+theta)/2 )*cos(-theta/2)`

`cos(-alpha)=cosalpha`, `phi=theta/2rArr2cos((2x+theta)/2 )=cos(x+φ)`

then:

`cos(x)+cos(x+θ)=Acos(x+φ)`

If `A=1/4rArr1/4=2cos(theta/2)`
`cos(theta/2)=1/8rArr theta/2=+-arccos(1/8)+2kpi`
General solution: `theta=+-2arccos(1/8)+4kpi`

for `0<=theta<=2pi` the solution is: `theta=2arccos(1/8)`

If `A=-1rArr-1=2cos(theta/2)`
`cos(theta/2)=-1/2rArr theta/2=+-2pi/3+2kpi`
General solution: `theta=+-4pi/3+4kpi`

for `0<=theta<=2pi` the solution is: `theta=4pi/3`

Note that in both cases the period is `4pi`, then is only one solution is valid in `[0,2pi].`

Answer 2

Please see below.

Explanation:

As `cosA+cosB=2cos((A+B)/2)cos((A-B)/2)`

`cosx+cos(x+theta)=2cos((2x+theta)/2)cos((-theta)/2)`

= `2cos(x+theta/2)cos(theta/2)`

= `2cos(theta/2)(x+theta/2)`

= `Acos(theta+phi)`

where `A=2cos(theta/2)` and `phi=theta/2`

(a) if `A=1/4`, we have `2cos(theta/2)=1/4`

or `cos(theta/2)=1/8` and

hence `costheta=2cos^2(theta/2)-1=2xx(1/8)^2-1=-31/32`

and `theta=cos^(-1)(-31/32)=2.891` or `2.891+pi`

(b) if `A=-1`, we have `2cos(theta/2)=-1`

or `cos(theta/2)=-1/2` and

hence `costheta=2cos^2(theta/2)-1=2xx(-1/2)^2-1=-1/2`

and `theta=cos^(-1)(-1/2)=(2pi)/3` or `(5pi)/3`

Answer 3

`theta=(4pi)/3 and` note that: `theta!=(2pi)/3ortheta!=(5pi)/3`
First part is already proved by Mr.Shwetank Mauria and Mr. Dhilak.

Explanation:

First part is proved .So we determine `theta` only for `A=-1`
Now,
`A=-1=>2cos(theta/2)=-1=>cos(theta/2)=-1/2`
We have,
`0<=theta<2pi to`, (given)
`=>0 <= (theta)/2 < pi=> 1^(st)and 2^(nd)` quadrant
But , `cos(theta/2)=-1/2<0=>2^(nd)`quadrant
So,
`cos(theta/2)=cos((2pi)/3)=>theta/2=(2pi)/3=>theta=(4pi)/3`
Note:
`A=2cos(theta/2)=2cos((2pi)/3)=2(-1/2)=-1`
For, `theta=(2pi)/3`
`=>color(red)A=2cos(theta/2)=2cos(((2pi)/3)/2)=2cos(pi/3)=2(1/2)=color(red)1`
`theta=(5pi)/3=>color(red)A=2cos(theta/2)=2cos(((5pi)/3)/2)=2cos((5pi)/6)=2cos(pi-pi/6)=-2cos(pi/6)=-2(sqrt3/2)=color(red)(-sqrt3)`