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

Solución

verificar

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

Verdadero

Pasos de solución

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

Manipular el lado derecho

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

Re-escribir usando identidades trigonométricas

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

Utilizar la identidad trigonométrica del ángulo doble:

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

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

Simplificar

(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)

Aplicar la formula del binomio al cuadrado:

(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)

Sumar elementos similares:

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

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

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

Utilizar la identidad pitagórica:

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

= 1

= 1

Se demostró que ambos lados pueden tomar la misma forma

\Rightarrow Verdadero

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)