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

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

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

prove

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

True

Solution steps

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

Manipulating right side

1 - sin (2 t)

Use the Pythagorean identity:

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

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

Use the Double Angle identity:

sin (2 t) = 2 sin (t) cos (t)

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

Apply Perfect Square Formula:

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

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

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

We showed that the two sides could take the same form

\Rightarrow True

(1 - sin (θ)) (1 + sin (θ)) = \frac{1}{sec ^{2} ( θ )}

sin (θ) = \frac{2 tan ( \frac{θ}{2} )}{1 + tan ^{2} ( \frac{θ}{2} )}

sin^{2} (x) - 2 cos^{2} (x) = 1 - 3 cos^{2} (x)

\frac{1}{sin ^{2} ( x )} = csc^{2} (x)

\frac{sin ( x )}{1 + cos ( x )} + cot (x) = csc (x)

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