Find the #cos(y - x)# if #cos(x) = 15/17#, #cot(y) = 24/7#, #0^@ < x < 90^@# and #90^@ < y < 180^@#?

1 Answer
Apr 2, 2018

#cos(x-y) = -304/425#

Explanation:

Start with the identity:

#cos(x-y) = cos(y)cos(x)+sin(y)sin(x)" [1]"#

We are given that #cos(x) = 15/17#

#cos(x-y) = cos(y)15/17+sin(y)sin(x)" [1.1]"#

Use the identity

#sin(x) = +-sqrt(1-cos^2(x))#

We are told that #0^@ < x < 90^@#, therefore, we shall use the positive case:

#sin(x) = sqrt(1-cos^2(x))#

Substitute #cos^2(x) = (15/17)^2#

#sin(x) = sqrt(1-(15/17)^2)#

#sin(x) = sqrt(289/289-225/289)#

#sin(x) = sqrt(64/289)#

#sin(x) = 8/17" [2]"#

Substitute equation [2] into equation [1.1]:

#cos(x-y) = cos(y)15/17+sin(y)8/17" [1.2]"#

Use the identity

#1+cot^2(y) = csc^2(y)#

The must be an error because y is the second quadrant but the cotangent is positive; I shall assume that it is negative. Substitute #cot^2(y) = (-24/7)^2#

#1+ (-24/7)^2=csc^2(y)#

#49/49+ 576/49=csc^2(y)#

#625/49=csc^2(y)#

Substitute #csc^2(y) = 1/sin^2(y)#

#625/49=1/sin^2(y)#

#sin^2(y) = 49/625#

#sin(y) = +-7/25#

Because we are told that #90^@ < y < 180^@#, we shall choose the positive value:

#sin(y) = 7/25" [3]"#

Substitute equation [3] into equation [1.2]:

#cos(x-y) = cos(y)15/17+7/25(8/17)" [1.2]"#

Use the identity:

#cot(y) = cos(y)/sin(y)#

Again assuming that the cotangent is negative:

#-24/7 = cos(y)/(7/25)#

#cos(y) = -24/7 7/25#

#cos(y) = -24/25" [4]"#

Substitute equation [4] into equation [1.2]:

#cos(x-y) = -24/25 (15/17)+7/25 (8/17)#

#cos(x-y) = -304/425#