Question

(sinx-cosx)squared=1+2 sinxcosx

Answer 1

1-2sinxcosx = 1+2sinxcosx
-sinxcosx = sinxcosx
sinxcosx = 0
\[\sin(x) = 0 or \cos(x) = 0\]
x=1/2n*pi, \[n \in \mathbb{R} \]

Answer 2

how did u do it from the beginning? should this become an identity or not?

Answer 3

? this is not an identity.
To simplify this
let sin x = a; cos x = b;
a^2+b^2 = 1;
(a-b)^2 = a^2+b^2 - 2ab

Answer 4

so for sinx and cosx, when(sinx-cosx)^2 is used, how did u use 1 and 2?

Answer 5

let sin x = a; cos x = b;
(a-b)^2
= (a-b)*(a-b)
= a*a - a*b - b*a + b*b
= a^2+b^2 - 2ab
a^2+b^2 = 1 (sin^2 x + cos^2 x = 1)
= 1 - 2ab

Answer 6

oh i get it! thanks! can u help me with the other question i posted up a few minutes ago?

Answer 7

its cosx/secxsinx=cscx-sinx

Answer 8

This is trig and you need to find x, let me try to help you on this one!