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

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

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

prove

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

True

Solution steps

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

Manipulating left side

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

Rewrite using trig identities

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

Use the Double Angle identity:

sin (2 x) = 2 sin (x) cos (x)

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

Simplify

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

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

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

Apply Perfect Square Formula:

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

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

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

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

Add similar elements:

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

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

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

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

sin (\frac{a}{2}) = \frac{1}{2} sin (a)

tan^{2} (α) = \frac{sin ^{2} ( α )}{1 - sin ^{2} ( α )}

tan (x) sec (x) (csc (x) - sin (x)) = 1

csc (x) cos (x) cot (x) = sin (x)

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