Is (cos(θ/2)-sin(θ/2))^2=1-sin(θ) ?

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

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prove (cos(θ/2)-sin(θ/2))^2=1-sin(θ)

prove

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

True

Solution steps

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

Let:

u = \frac{θ}{2}

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

Prove

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

True

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

Manipulating left side

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

Expand

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

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

Apply Perfect Square Formula:

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

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

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

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

Rewrite using trig identities

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

Use the Double Angle identity:

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

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

Use the Pythagorean identity:

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

= 1 - sin (2 u)

= 1 - sin (2 u)

We showed that the two sides could take the same form

\Rightarrow True

Therefore

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

We showed that the two sides could take the same form

\Rightarrow True

1 + sin (4 a) + sin (2 a) = - 2 cos (a)

(csc^{2} (θ)) (tan^{2} (θ)) - 1 = tan^{2} (θ)

\frac{csc ( θ ) + 1}{1 + sin ( θ )} = csc (θ)

sec (A) csc (A) = csc (A) tan (A)

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

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