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
A)
B)
C)
D)
THE ANSWER IS A FOR YOUR REFERENCE
3 Answers
As
So
See below.
Explanation:
Using de Moivre's identity
now multiplying both equations
and then
so neither of the options is contemplated.
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 2A = 2cos^2A -1 #
If we replace
# cos B = 2cos^2(B/2) -1 => 2cos^2(B/2) =1+cosB#
# :. cos^2(B/2) =1/2+cosB/2#
So then, putting
# 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
# cos((alpha-beta)/2) = -3/sqrt(130) #