Vtyjkyuk

Find the value of $alpha,beta$ for the equation $cosalpha cosbeta cos(alpha +beta)=-frac18$

$alpha>0$ & $beta<fracpi2$

I get the following step after some substitution

$cos2alpha + cos2beta+cos2(alpha+beta)=-frac32$ from here not able to proceed.

Suppose $$0<alpha,beta<fracpi2.tag1$$ From
$$cos2alpha + cos2beta+cos2(alpha+beta)=-frac32$$
one has
$$ cos^2alpha+cos^2beta+cos^2(alpha+beta)=frac34. $$
Note that
$$cosalpha cosbeta cos(alpha +beta)=-frac18tag2$$
implies
$$cos^2alpha cos^2beta cos^2(alpha +beta)=frac164.$$
By the AM-GM inequality
$$ a+b+cge3sqrt[3]abc$$
one has
$$ frac34=cos^2alpha+cos^2beta+cos^2(alpha+beta)=3sqrt[3]cos^2alphacos^2betacos^2(alpha+beta)=3sqrt[3]frac164=frac34 $$
and the equal sign holds if and only if
$$cos^2alpha=cos^2beta=cos^2(alpha+beta).tag3$$
From (1)(2)(3), it is easy to see
$$ alpha=beta=fracpi3. $$

One has $$cos alphacosbeta(cosalphacosbeta - sinalphasinbeta) = -frac18$$
$$1 - tanalphatanbeta = -frac18(1+tan^2alpha)(1+tan^2beta)$$
$$8-8tanalphatanbeta = -1-tan^2alpha-tan^2beta - tan^2alphatan^2beta$$
$$(tanalphatanbeta-3)^2 + (tanalpha-tanbeta)^2 = 0$$
One then has
$$tanalpha=tanbeta=sqrt3.$$

Like $ cos A cos B cos C leq frac18 $,

$$-1=8cosalphacosbetacos(alpha+beta)=4[cos(alpha-beta)+cos(alpha+beta)]cos(alpha+beta)$$

$$iffcos^2(alpha+beta)+cos(alpha+beta)cos(alpha-beta)+dfrac14=0$$

which is a Quadratic Equation in $cos(alpha+beta)$ which is real,

so, the discriminant must be $ge0$

i.e., $$0lecos^2(alpha-beta)-1=-sin^2(alpha-beta)$$

$$implies(i)sin(alpha-beta)=0$$

$impliesalpha-beta=mpi$ where $m$ is any integer

As $0<alpha,beta<dfracpi2,m=0impliescos(alpha-beta)=1$

and $$(ii)cos(alpha+beta)=-dfraccos(alpha-beta)2impliescos2alpha=dfrac12$$