# 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)

Mar 16, 2018

If $A = \frac{1}{4}$

for $0 \le \theta \le 2 \pi$ the solution is: $\theta = 2 \arccos \left(\frac{1}{8}\right)$

If $A = - 1$
for $0 \le \theta \le 2 \pi$ the solution is: $\theta = 4 \frac{\pi}{3}$

#### Explanation:

Using prosthaphaeresis formula:
$\cos \alpha + \cos \beta = 2 \cos \frac{\alpha + \beta}{2} \cdot \cos \frac{\alpha - \beta}{2}$
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)

$\cos \left(- \alpha\right) = \cos \alpha$, phi=theta/2rArr2cos((2x+theta)/2 )=cos(x+φ)

then:

cos(x)+cos(x+θ)=Acos(x+φ)

If $A = \frac{1}{4} \Rightarrow \frac{1}{4} = 2 \cos \left(\frac{\theta}{2}\right)$
$\cos \left(\frac{\theta}{2}\right) = \frac{1}{8} \Rightarrow \frac{\theta}{2} = \pm \arccos \left(\frac{1}{8}\right) + 2 k \pi$
General solution: $\theta = \pm 2 \arccos \left(\frac{1}{8}\right) + 4 k \pi$

for $0 \le \theta \le 2 \pi$ the solution is: $\theta = 2 \arccos \left(\frac{1}{8}\right)$

If $A = - 1 \Rightarrow - 1 = 2 \cos \left(\frac{\theta}{2}\right)$
$\cos \left(\frac{\theta}{2}\right) = - \frac{1}{2} \Rightarrow \frac{\theta}{2} = \pm 2 \frac{\pi}{3} + 2 k \pi$
General solution: $\theta = \pm 4 \frac{\pi}{3} + 4 k \pi$

for $0 \le \theta \le 2 \pi$ the solution is: $\theta = 4 \frac{\pi}{3}$

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

Mar 16, 2018

#### Explanation:

As $\cos A + \cos B = 2 \cos \left(\frac{A + B}{2}\right) \cos \left(\frac{A - B}{2}\right)$

$\cos x + \cos \left(x + \theta\right) = 2 \cos \left(\frac{2 x + \theta}{2}\right) \cos \left(\frac{- \theta}{2}\right)$

= $2 \cos \left(x + \frac{\theta}{2}\right) \cos \left(\frac{\theta}{2}\right)$

= $2 \cos \left(\frac{\theta}{2}\right) \left(x + \frac{\theta}{2}\right)$

= $A \cos \left(\theta + \phi\right)$

where $A = 2 \cos \left(\frac{\theta}{2}\right)$ and $\phi = \frac{\theta}{2}$

(a) if $A = \frac{1}{4}$, we have $2 \cos \left(\frac{\theta}{2}\right) = \frac{1}{4}$

or $\cos \left(\frac{\theta}{2}\right) = \frac{1}{8}$ and

hence $\cos \theta = 2 {\cos}^{2} \left(\frac{\theta}{2}\right) - 1 = 2 \times {\left(\frac{1}{8}\right)}^{2} - 1 = - \frac{31}{32}$

and $\theta = {\cos}^{- 1} \left(- \frac{31}{32}\right) = 2.891$ or $2.891 + \pi$

(b) if $A = - 1$, we have $2 \cos \left(\frac{\theta}{2}\right) = - 1$

or $\cos \left(\frac{\theta}{2}\right) = - \frac{1}{2}$ and

hence $\cos \theta = 2 {\cos}^{2} \left(\frac{\theta}{2}\right) - 1 = 2 \times {\left(- \frac{1}{2}\right)}^{2} - 1 = - \frac{1}{2}$

and $\theta = {\cos}^{- 1} \left(- \frac{1}{2}\right) = \frac{2 \pi}{3}$ or $\frac{5 \pi}{3}$

Mar 16, 2018

$\theta = \frac{4 \pi}{3} \mathmr{and}$ note that: $\theta \ne \frac{2 \pi}{3} \mathmr{and} \theta \ne \frac{5 \pi}{3}$
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,
$A = - 1 \implies 2 \cos \left(\frac{\theta}{2}\right) = - 1 \implies \cos \left(\frac{\theta}{2}\right) = - \frac{1}{2}$
We have,
$0 \le \theta < 2 \pi \to$, (given)
$\implies 0 \le \frac{\theta}{2} < \pi \implies {1}^{s t} \mathmr{and} {2}^{n d}$ quadrant
But , $\cos \left(\frac{\theta}{2}\right) = - \frac{1}{2} < 0 \implies {2}^{n d}$quadrant
So,
$\cos \left(\frac{\theta}{2}\right) = \cos \left(\frac{2 \pi}{3}\right) \implies \frac{\theta}{2} = \frac{2 \pi}{3} \implies \theta = \frac{4 \pi}{3}$
Note:
$A = 2 \cos \left(\frac{\theta}{2}\right) = 2 \cos \left(\frac{2 \pi}{3}\right) = 2 \left(- \frac{1}{2}\right) = - 1$
For, $\theta = \frac{2 \pi}{3}$
$\implies \textcolor{red}{A} = 2 \cos \left(\frac{\theta}{2}\right) = 2 \cos \left(\frac{\frac{2 \pi}{3}}{2}\right) = 2 \cos \left(\frac{\pi}{3}\right) = 2 \left(\frac{1}{2}\right) = \textcolor{red}{1}$
$\theta = \frac{5 \pi}{3} \implies \textcolor{red}{A} = 2 \cos \left(\frac{\theta}{2}\right) = 2 \cos \left(\frac{\frac{5 \pi}{3}}{2}\right) = 2 \cos \left(\frac{5 \pi}{6}\right) = 2 \cos \left(\pi - \frac{\pi}{6}\right) = - 2 \cos \left(\frac{\pi}{6}\right) = - 2 \left(\frac{\sqrt{3}}{2}\right) = \textcolor{red}{- \sqrt{3}}$