Asked by Phum

SinA+ cosA=17/3 find sinA-cosA
Answered by Reiny
sinA + cosA = 17/3
square both sides
sin^2 A + 2sinAcosA + cos^2 = 289/9
1 + 2sinAcosA = 289/9
2sinAcosA = 280/9
sinAcosA = 140/9

let x = sinA - cosA
x^2 = sin^2 A - 2sinAcosA + cos^2 A
= 1 - 2sinAcosA
= 1 - 280/9
= - 271/9
which is not possible, (x^2 cannot be negative)

Let's think about this.
y = sinA + cosA is a sinusoidal function with an amplitude of √(1^2 + 1^2) = √2
which is less than 17/3

sinA + cosA = 17/3 is not possible, I suspect a typo.
Answered by Phum
SinA+SinA=17/3
Answered by Steve
still not possible.

sinA cannot be greater than 1.

17/3 > 2, so sinA+sinA can never be 17/3, which is almost 6!

Better revisit the original problem.
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