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Let
y = sin(x) + cos(x)
Since x is in quadrant Q1, this means both sine and cosine are positive here.
Overall y = sin(x) + cos(x) is also positive.
Square both sides of the original equation.
sin(x) - cos(x) = 1/3
[ sin(x) - cos(x) ]^2 = (1/3)^2
sin^2(x) - 2sin(x)cos(x) + cos^2(x) = 1/9
sin^2(x) + cos^2(x) - 2sin(x)cos(x) = 1/9
1 - 2sin(x)cos(x) = 1/9
-2sin(x)cos(x) = 1/9-1
-2sin(x)cos(x) = 1/9-9/9
-2sin(x)cos(x) = -8/9
2sin(x)cos(x) = 8/9
We'll use this later.
Now let's square both sides of the y equation we set up earlier.
y = sin(x) + cos(x)
y^2 = [ sin(x) + cos(x) ]^2
y^2 = sin^2(x) + 2sin(x)cos(x) + cos^2(x)
y^2 = sin^2(x) + cos^2(x) + 2sin(x)cos(x)
y^2 = 1 + 2sin(x)cos(x)
We'll replace the "2sin(x)cos(x)" with 8/9 since we found it earlier in the section above.
y^2 = 1 + 2sin(x)cos(x)
y^2 = 1 + 8/9
At this point, the goal is to solve for y to get the result of sin(x)+cos(x).
y^2 = 1 + 8/9
y^2 = 9/9 + 8/9
y^2 = (9 + 8)/9
y^2 = 17/9
y = sqrt(17/9) ..... y is positive
y = sqrt(17)/sqrt(9)
y = sqrt(17)/3
Once again, y is positive so that's why we ignored the minus when going from y^2 = 17/9 to y = sqrt(17)/3.
Answer: Choice B. sqrt(17)/3
This can be written as
The "3" is not inside the square root.
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