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

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

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

prove

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

True

Solution steps

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

Manipulating left side

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

Simplify

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

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

(cos (x) + sin (x))^{2} : cos^{2} (x) + 2 cos (x) sin (x) + sin^{2} (x)

Apply Perfect Square Formula:

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

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

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

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

Add similar elements:

2 cos (x) sin (x) - 2 cos (x) sin (x) = 0

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

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

Rewrite using trig identities

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

Use the Pythagorean identity:

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

= 1

= 1

We showed that the two sides could take the same form

\Rightarrow True

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

(1 - sin (3 a)) (sin (3 a) + 1) = cos^{2} (3 a)

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

cos^{2} (\frac{( 12 π )}{2}) = cos^{2} (12 π)

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

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