Is (sin(x)-cos(x))^2-1=-2sin(x)cos(x) ?

The answer to whether (sin(x)-cos(x))^2-1=-2sin(x)cos(x) is True

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prove (sin(x)-cos(x))^2-1=-2sin(x)cos(x)

prove

(sin (x) - cos (x))^{2} - 1 = - 2 sin (x) cos (x)

True

Solution steps

(sin (x) - cos (x))^{2} - 1 = - 2 sin (x) cos (x)

Manipulating left side

(sin (x) - cos (x))^{2} - 1

Rewrite using trig identities

(sin (x) - cos (x))^{2} - 1

Use the Pythagorean identity:

1 = cos^{2} (x) + sin^{2} (x)

= (sin (x) - cos (x))^{2} - cos^{2} (x) + sin^{2} (x)

Simplify

(sin (x) - cos (x))^{2} - cos^{2} (x) + sin^{2} (x) : 2 sin^{2} (x) - 2 sin (x) cos (x)

(sin (x) - cos (x))^{2} - cos^{2} (x) + sin^{2} (x)

(sin (x) - cos (x))^{2} : sin^{2} (x) - 2 sin (x) cos (x) + cos^{2} (x)

Apply Perfect Square Formula:

(a - b)^{2} = a^{2} - 2 ab + b^{2}

a = sin (x), b = cos (x)

= sin^{2} (x) - 2 sin (x) cos (x) + cos^{2} (x)

= sin^{2} (x) - 2 sin (x) cos (x) + cos^{2} (x) - cos^{2} (x) + sin^{2} (x)

Simplify

sin^{2} (x) - 2 sin (x) cos (x) + cos^{2} (x) - cos^{2} (x) + sin^{2} (x) : 2 sin^{2} (x) - 2 sin (x) cos (x)

sin^{2} (x) - 2 sin (x) cos (x) + cos^{2} (x) - cos^{2} (x) + sin^{2} (x)

Add similar elements:

cos^{2} (x) - cos^{2} (x) = 0

= sin^{2} (x) - 2 sin (x) cos (x) + sin^{2} (x)

Add similar elements:

sin^{2} (x) + sin^{2} (x) = 2 sin^{2} (x)

= 2 sin^{2} (x) - 2 sin (x) cos (x)

= 2 sin^{2} (x) - 2 sin (x) cos (x)

= 2 sin^{2} (x) - 2 sin (x) cos (x)

= 2 sin^{2} (x) - 2 sin (x) cos (x)

= - 2 sin (x) cos (x)

We showed that the two sides could take the same form

\Rightarrow True

- 2 tan^{2} (0) = 2 tan (0)

2 cot (x) sin (x) = 2 cos (x)

csc^{2} (x) (sin^{2} (x) - 1) = - cot^{2} (x)

cos (3 x) = cos (x) (4 cos^{2} (x) - 3)

\frac{1}{sin ( x ) cot ( x )} = sec (x)

Is (sin(x)-cos(x))^2-1=-2sin(x)cos(x) ?
The answer to whether (sin(x)-cos(x))^2-1=-2sin(x)cos(x) is True