SOLUTION: Given cos(x) = 1&#8260;2 with 0o < x < 90o, and cos(y) = 1&#8260;7 with 270o < y < 360o. Find cos(x + y).

Algebra ->  Trigonometry-basics -> SOLUTION: Given cos(x) = 1&#8260;2 with 0o < x < 90o, and cos(y) = 1&#8260;7 with 270o < y < 360o. Find cos(x + y).      Log On


   

Question 811262: Given cos(x) = 1⁄2 with 0o < x < 90o, and cos(y) = 1⁄7 with 270o < y < 360o. Find cos(x + y).
Answer by lwsshak3(11628) About Me  (Show Source):
You can put this solution on YOUR website!
Given cos(x) = 1⁄2 with 0o < x < 90o, and cos(y) = 1⁄7 with 270o < y < 360o. Find exact value of cos(x + y).
***
cos(x)=1/2(Q1)
sin(x)=√(1-cos^(x))=√(1-1/4)=√(3/4)=√3/2
cos(y)=1/7(Q4)
sin(y)=-√(1-cos^(y))=-√(1-1/49)=-√(48/49)=-√48/7=-4√3/7
cos(x+y)=cos(x)cos(y)-sin(x)sin)y)
=(1/2)*(1/7)+√3/2*4√3/7
=1/14+12/14=13/14
..
Check:(w/calculator)
cos(x)=1/2
x=60˚
cos(y)=1/7
y≈278.21˚
x+y≈338.21˚
..
cos(x+y)≈cos(338.21˚)≈0.9285..
exact value=13/14≈0.9285..