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

Solución

verificar

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

Verdadero

Pasos de solución

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

Manipular el lado derecho

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

Simplificar

(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)

Aplicar la formula del binomio al cuadrado:

(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)

Sumar elementos similares:

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

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

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

Re-escribir usando identidades trigonométricas

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

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)