Given that: sin α = 15/17, with α in QI, and sin(α + β) = 12/13, with α + β in QII, find sin β and cos β?

Given that:
sin α = 15/17, with α in QI, and sin(α + β) = 12/13, with α + β in QII, find sin β and cos β

1 Answer
Feb 10, 2018

# cos beta = 140/221 \ \ # and # \ \ sin beta= 171/221 #

Explanation:

Using #sin^2A+cos^2A -= 1# we can write:

# cos^2 alpha =1 - sin^2 alpha #
# \ \ \ \ \ \ \ \ \ = 1-(15/17)^2 #
# \ \ \ \ \ \ \ \ \ = 1-225/289 #
# \ \ \ \ \ \ \ \ \ = 64/289 #
# :. cos alpha = sqrt(64/289) = +- 8/17 #

Knowing that #alpha# is acute we can discard the negative solution leaving us with:

# cos alpha=8/17 # ..... [A]

Similarly, we know that:

# cos^2 (alpha+beta) =1 - sin^2 (alpha+beta) #
# \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ =1 - (12/13)^2 #
# \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ =1 - 144/169 #
# \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ =25/169 #
# :. cos (alpha+beta) = sqrt(25/169) = +- 5/13 #

Knowing that #alpha+beta# is obtuse we can discard the positive solution leaving us with:

# cos(alpha+beta) = -5/13 # ..... [B]

Now, we can use the sum of angle formula:

# sin(A+B) = sinAcosB+cosAsinB #
# cos(A+B) = cosAcosB-sinAsinB #

From which we get, using [A] and [B], that:

# sin(alpha+beta)= sin alpha cos beta+cos alpha sin beta #
# :. 12/13= 15/17 cos beta + 8/17 sin beta # ..... [C]

And:

# cos(alpha+beta)= cos alpha cos beta - sin alpha sin beta #
# :. -5/13 = 8/17 cos beta - 15/17 sin beta # ..... [D]

We now have two simultaneous equations [C] and [D] in #sin beta# and #cos beta#, which can solve to get:

# cos beta = 140/221 \ \ # and # \ \ sin beta= 171/221 #