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

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

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

prove

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

True

Solution steps

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

Manipulating left side

(cos (x) + sin (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

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

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

8 csc^{2} (x) - 3 cot^{2} (x) = 3 + 5 csc^{2} (x)

sin^{2} (t) = (sin (t))^{2}

sin (2 x) - 2 cos (x) = 0, (0, 2 π)

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