Yahoo Answers: Answers and Comments for If cos x +cos y=1, sec x+sec y=4, where 0°<x<180°, find value of x+y.? [Mathematics]
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From Anonymous
enUS
Sun, 19 May 2019 16:10:55 +0000
3
Yahoo Answers: Answers and Comments for If cos x +cos y=1, sec x+sec y=4, where 0°<x<180°, find value of x+y.? [Mathematics]
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https://answers.yahoo.com/question/index?qid=20190519161055AAaUuUb
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From Mewtwo: Using the second equation:
4 = sec x + sec y ...
https://answers.yahoo.com/question/index?qid=20190519161055AAaUuUb
https://answers.yahoo.com/question/index?qid=20190519161055AAaUuUb
Sun, 19 May 2019 19:42:49 +0000
Using the second equation:
4 = sec x + sec y
= 1 / cos x + 1 / cos y
= (cos y + cos x) / (cos x cos y)
= 1 / (cos x cos y), or
cos x cos y = 1/4.
Using the first equation, cos y = 1  cos x. Plugging this into the equivalent form of the second equation that we found, then,
cos x(1  cos x) = 1/4, or
4 cos x(1  cos x) = 1.
Thus,
4 cos x  4 cos^2 x = 1
4 cos^2 x  4 cos x + 1 = 0
Factoring this gives
(2 cos x  1)^2 = 0.
Thus,
2 cos x  1 = 0
cos x = 1/2.
Since 0° < x < 180°, we verify then that x = 60° (Use your unit circle or a calculator).
Thus,
cos y = 1  cos x
= 1  cos 60°
= 1  1/2
= 1/2.
So, y = 60° as well. Therefore, x + y = 60° + 60° = 120°.