SinX +Cos X = K? TRIG HELP

If sinX+cosX = K, what values of k does SinXCosX=1 ????

Square both sides:

sin(x)^2 + 2sin(x)cos(x) + cos(x)^2 = k^2
1 + 2sin(x)cos(x) = k^2
2 * sin(x)cos(x) = k^2 - 1
sin(x)cos(x) = (1/2) * (k^2 - 1)
sin(x)cos(x) = 1
1 = (1/2) * (k^2 - 1)
2 = k^2 - 1
3 = k^2
k = -sqrt(3) , sqrt(3)

It's probably wrong.

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None. For all x, |sin x| <= 1 and |cos x| <=1, and they are never both 1 in magnitude. You can also show this using the sine double-angle identity:

sin (2x) = 2(sin x)(cos x) ... true for all x

If (sin x)(cos x) were equal to 1 for some value of x, then sin (2x) would be 2 for that value of x. That's impossible, so no such x exists.

Hi.

This is the procedure:

sinXcosX=1 <---- this is the key. as you know sinX is betweeen -1 and 1, same with cosX.

So, sinXcosX is possible only if sinX = 1 and cosX =1 or if sinX = -1 and cosX = -1, and there is no value of X that meet that requirements.

So the solution is = none.

There is no real solution to this problem
Note that sin x cos x = 1 gives
2 sin x cos x = 2
or sin 2x = 2.
But -1 <= sin 2x <= 1
so there is no real solution to this problem.

sinx + cosx = K

sinxcosx = 1
This implies that sinx and cosx are reciprocals, which is not possible.
No real solution.