Question

Let `alpha,beta` be such that `pi < alpha - beta < 3pi`. If `sinalpha + sinbeta = -21/65` and`cosalpha + cosbeta = -27/65`, then the value of `cos(alpha - beta)/2` is?

A) `-30/(sqrt130)`
B) `3/(sqrt130)`
C) `6/65`
D) `-6/65`
THE ANSWER IS A FOR YOUR REFERENCE

Answer 1

`(cosalpha+cosbeta)^2+(sinalpha+sinbeta)^2=(-27/65)^2+(-21/65)^2`

`=>2+2(cosalphacosbeta+sinalphasinbeta)=1170/65^2`

`=>2(1+cos(alpha-beta))=1170/65^2`

`=>4cos^2((alpha-beta)/2)=1170/65^2`

`=>2cos((alpha-beta)/2)=-sqrt1170/65`

`=>cos((alpha-beta)/2)=-sqrt1170/130`

`=-sqrt(9*130)/130=-3/sqrt130`

As `pi<(alpha-beta)<3pi`

`=>pi/2<(alpha-beta)/2<(3pi)/2`

`(alpha-beta)/2` is in 3rd quadrant.

So `cos((alpha-beta)/2)` is negative

Answer 2

See below.

Explanation:

Using de Moivre's identity

`e^(ix) = cos x + i sinx`

`e^(ialpha)+ e^(i beta) = z_0` and

`e^(-ialpha)+ e^(-i beta) = bar z_0`

now multiplying both equations

`(e^(ialpha)+ e^(i beta))(e^(-ialpha)+ e^(-i beta)) = z_0 bar z_0` or

`2+ e^(i(alpha-beta))+e^(-i(alpha-beta)) = norm(z_0)^2` or

`(e^(i(alpha-beta))+e^(-i(alpha-beta)))/2 = cos(alpha-beta) = (norm(z_0)^2-2)/2 = ((21/65)^2+(27/65)^2-2)/2=-56/65`

and then

`(cos(alpha-beta))/2 = -28/65`

so neither of the options is contemplated.

Answer 3

If we look at:

`cos(alpha - beta)/2 = -28/65`

If, however, we look at:

`cos((alpha - beta)/2) = -3/sqrt(130)`

In either case, None of the quoted answers are correct.

Explanation:

We are given that:

`sin alpha+sin beta \ = -21/65` ..... [A]
`cos alpha+cos beta = -27/65` ..... [B]

Squaring we:

`[A] => (sin alpha+sin beta)^2 = (-21/65)^2`
`:. sin^2 alpha + 2sin alpha sin beta + sin^2 beta = (21/65)^2` ..... [C]

`[B] => (cos alpha+cos beta)^2 = (-27/65)^2`
`:. cos^2alpha + 2cosalphacosbeta + cos^2beta = (27/65)^2` ..... [D]

Adding [C]+{D] gives:

`sin^2 alpha + cos^2alpha + 2sin alpha sin beta + 2cosalphacosbeta + sin^2 beta + cos^2beta = (21/65)^2 +(27/65)^2`

`:. 1 + 2(sin alpha sin beta + cosalphacosbeta) + 1 = (21/65)^2 +(27/65)^2` ..... [E]

Now, the trigonometric sum/difference identity gives:

`cos(alpha - beta) = cos alpha \ cos beta + sin alpha \ sin beta`

And so inserting this result into [E] we get:

`1 + 2cos(alpha - beta) + 1 = (21/65)^2 +(27/65)^2`

`:. 2cos(alpha - beta) + 2 = (21^2+27^2)/65^2`

`:. 2cos(alpha - beta) = (21^2+27^2 -2*65^2)/65^2`

`:. 2cos(alpha - beta) = (441+729-8450)/4225`

`:. 2cos(alpha - beta) = -7280/4225`

`:. 2cos(alpha - beta) = -112/65`

`:. cos(alpha - beta)/2 = (-112/65)/4`

`:. cos(alpha - beta)/2 = -28/65`

If, however, you wanted to calculate `cos((alpha-beta)/2)`, then the calculation is as above, and in addition we use the identity

`cos 2A = 2cos^2A -1`

If we replace `A` by `B/2`, then we get:

`cos B = 2cos^2(B/2) -1 => 2cos^2(B/2) =1+cosB`
`:. cos^2(B/2) =1/2+cosB/2`

So then, putting `B=alpha-beta`, and using the result for `cos(alpha - beta)/2` we would get

`cos^2((alpha-beta)/2) =1/2-28/65`
`:. cos^2((alpha-beta)/2) = 9/130`
`:. cos((alpha-beta)/2) = +-3/sqrt(130)`

We are also given that:

`pi lt alpha - beta lt 3pi`
`:. pi/2 lt (alpha - beta)/2 lt 3/2pi`

If we look at the graph of `y=cosx`, then we see for `pi/2 lt x lt 3/2pi` that `y=cosx lt 0` and thus we can eliminate the positive solution, giving us

`cos((alpha-beta)/2) = -3/sqrt(130)`