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

3 Answers
Aug 18, 2017

#(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

Aug 18, 2017

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.

Aug 18, 2017

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 #

https://socratic.org/questions/how-do-you-find-the-stationary-points-of-the-function-y-sin-x

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) #