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

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

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

prove

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

True

Solution steps

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

Let:

u = \frac{x}{2}

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

Prove

1 - sin (2 u) = (sin (u) - cos (u))^{2} :

True

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

Manipulating left side

1 - sin (2 u)

Use the Pythagorean identity:

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

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

Use the Double Angle identity:

sin (2 u) = 2 sin (u) cos (u)

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

Apply Perfect Square Formula:

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

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

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

We showed that the two sides could take the same form

\Rightarrow True

Therefore

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

We showed that the two sides could take the same form

\Rightarrow True

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