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

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

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

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

prove

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

True

Solution steps

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

Manipulating left side

(cos (θ) - sin (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^{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)

We showed that the two sides could take the same form

\Rightarrow True

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