Prove the following trigonometric identities
1 answer:
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There are two solutions
The first solution is (x,y) = (3,0)
The second solution is (x,y) = (0,3)
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Work Shown:
Isolate x in the first equation (subtract y from both sides)
x+y = 3
x = -y+3
Now plug this into the second equation. Isolate y.
x^2 + y^2 = 9
(-y+3)^2 + y^2 = 9 ... replace x with -y+3
y^2 - 6y + 9 + y^2 = 9 ... FOIL rule
2y^2 - 6y + 9 = 9
2y^2 - 6y = 0
2y(y-3) = 0
2y = 0 or y-3 = 0 .... zero product property
y = 0 or y = 3
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If y = 0, then x is
x = -y+3
x = -0+3
x = 3
Therefore, x = 3 and y = 0 pair up to get the ordered pair (x,y) = (3,0) as one solution of this system
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If y = 3, then x is
x = -y+3
x = -3+3
x = 0
Therefore, x = 0 and y = 3 pair up to get the ordered pair (x,y) = (0,3) as the other solution of this system
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If you were to graph the two equations, you should find that the curves cross each other at (0,3) and (3,0). I've marked these two points as A and B respectively. The order of the solutions does not matter.
