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

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

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

prove

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

True

Solution steps

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

Manipulating left side

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

Expand

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

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

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

Rewrite using trig identities

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

Use the Pythagorean identity:

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

= 1 + 2 cos (x) sin (x)

= 1 + 2 cos (x) sin (x)

= 2 cos (x) sin (x) + 1

We showed that the two sides could take the same form

\Rightarrow True

\frac{cot ^{2} ( x )}{csc ( x ) + 1} + 1 = csc (x)

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

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

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

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

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