help! is (sinx)(cosx)=1?

I am not quite clear what you want, but I will try to be of help.

Forgive me if it is not what you were looking for

as your question got tailed off when posted.

Part I

If you mean is (sinx)(cosx) = 1 a valid identity (?) the answer is no.

If you mean is there an x value for which (sinx)(cosx) = 1

sin(x) is in -1 to +1 as is cos(x), so 1/cos(x) is outside -1 to +1

So they never intersect, which means there are no real values for x.

Another way to show this is to try to find solutions to

(sinx)^2*[1 - (sinx)^2] = 1

equivalent to finding solutions to

s^4 - s^2 + 1 = 0

This has no real solution

Part II

If you mean is 1-sinx = cosx a valid identity (?) the answer is no.

If you mean are there x values for which 1-sinx = cosx

Yes, as would be expected since both are infinitely continuous in -1 to +1

To solve sinx + cosx = 1 square both sides

[sin(x)]^2 + 2sinx cosx + [cos(x)]^2 = 1

but [sin(x)]^2 + [cos(x)]^2 = 1 so subtracting

2sinx cosx = 0 but this is the same as

sin 2x = 0

This has solutions 2x = 0 and 2x = pi and 2x = 2pi and so on

so x = 0, pi/2, pi, 3pi/2, 2pi...... etc

I hope that is the sort of thing you needed,

Regards - Ian

sin^2(x) + cos^2(x) = 1 so 1-sin^2(x) = cos^2(x)

search the web for for "trigonometric identities" and you should find a wealth of resources

..... if sin(x) cos(x) = 1

then: sin²(x) cos² (x) = 1

... or sin²(x) (1-sin²(x)) = 1

let u = sin²(x)

... u (1-u) = 1

... u-u² = 1

... 0 = u² - u + 1

determinant

= b² - 4ac = -5

= 2 imaginary roots

therefore there are no real solutions for u

and thus there are no real solutions for sin(x)