SinA+ cosA=17/3 find sinA-cosA

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.

To find the value of sinA - cosA, we need to use the given equation, sinA + cosA = 17/3, and manipulate it to obtain the desired expression.

Step 1: Square both sides of the equation sinA + cosA = 17/3.
(sinA + cosA)^2 = (17/3)^2
sin^2A + 2sinAcosA + cos^2A = 289/9

Step 2: Since sin^2A + cos^2A = 1 (a well-known trigonometric identity), we can substitute 1 in place of sin^2A + cos^2A in the equation obtained from Step 1.
1 + 2sinAcosA = 289/9

Step 3: Rearrange the equation to isolate sinAcosA.
2sinAcosA = 289/9 - 1
2sinAcosA = 289/9 - 9/9
2sinAcosA = 280/9

Step 4: Simplify the equation further by dividing both sides by 2.
sinAcosA = 280/9 / 2
sinAcosA = 140/9

Now, we have the value of sinAcosA. To find sinA - cosA, we'll need to make use of an additional identity:

sin(A - B) = sinAcosB - cosAsinB

In this case, we can write sinA - cosA as sin(A - A). Hence,

sin(A - A) = sinAcosA - cosAsinA

Since sin(A - A) is equivalent to sin0, which is 0, we substitute 0 for sin(A - A) and sinAcosA for sinAcosA in the equation:

0 = sinAcosA - cosAsinA

Rearranging this equation, we get:

- sinA = - cosAsinA

Dividing both sides by -1, we obtain:

sinA = cosAsinA

Now we can substitute sinA = cosAsinA in the equation sinA + cosA = 17/3:

cosAsinA + cosA = 17/3

Factoring out cosA, we have:

cosA(sinA + 1) = 17/3

Finally, we can find the value of sinA - cosA by substituting sinA = cosAsinA in the above equation:

cosA(cosAsinA + 1) = 17/3

cosA(cosA * cosA + 1) = 17/3

cosA(cos^2A + 1) = 17/3

Since we have only one equation with two variables (cosA and cos^2A), we can't directly solve for the value of sinA - cosA without any further information.