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

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

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

prove

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

True

Solution steps

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

Manipulating left side

(sin (θ) - cos (θ))^{2}

Expand

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

(- cos (θ) + sin (θ))^{2}

Apply Perfect Square Formula:

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

a = - cos (θ), b = sin (θ)

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

Simplify

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

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

Remove parentheses:

(- a) = - a

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

Apply exponent rule:

(- a)^{n} = a^{n},

n

is even

(- cos (θ))^{2} = cos^{2} (θ)

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

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

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

Rewrite using trig identities

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

Use the Pythagorean identity:

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

= 1 - 2 cos (θ) sin (θ)

= 1 - 2 cos (θ) sin (θ)

= 1 - 2 sin (θ) cos (θ)

We showed that the two sides could take the same form

\Rightarrow True

cot (θ) sin (θ) sec (θ) = 1

cos^{4} (x) = 1 - 2 sin^{2} (x) + sin^{4} (x)

tan (x) \cdot cos (x) = sin (x)

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

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

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