SinA+ cosA=17/3 find sinA-cosA

3 answers

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.
SinA+SinA=17/3
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.
Similar Questions
  1. Hi, can someone give me the answer to this question:secA-tanA=1/4. Find cosA I got: CosA=4(1-sinA) Is my answer good? Now
    1. answers icon 2 answers
  2. Hello I am having trouble with proving this:sinA/secA+secA/cscA = tan(A)(2-sin^2(A)) So far I've gotten SinA/(1/cosA) +
    1. answers icon 1 answer
    1. answers icon 1 answer
  3. Let P=(cosA sinA)(sinA cosA) Q=(cosB sinA) (sinB cosA) show that PQ=(cosA(A-B) sin(A B)) (sinA(A B) cos(A-B))
    1. answers icon 0 answers
more similar questions