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)

3 Answers
Mar 16, 2018

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].

Mar 16, 2018

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

Mar 16, 2018

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)