Is (sin(θ)-cos(θ))^2=1-sin(2θ) ?

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

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prove (sin(θ)-cos(θ))^2=1-sin(2θ)

prove

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

True

Solution steps

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

Manipulating right side

1 - sin (2 θ)

Use the Pythagorean identity:

1 = cos^{2} (θ) + sin^{2} (θ)

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

Use the Double Angle identity:

sin (2 θ) = 2 sin (θ) cos (θ)

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

Apply Perfect Square Formula:

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

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

= (sin (θ) - cos (θ))^{2}

We showed that the two sides could take the same form

\Rightarrow True

\frac{sec ( x )}{csc ( x )} = tan (x)

\frac{csc ( x )}{sec ( x )} = cot (x)

cot^{2} (θ) (1 + tan^{2} (θ)) = csc^{2} (θ)

sin (2 A) = 2 sin (A)

tan (θ) cos (θ) csc (θ) = 1

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