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verificar (1-cos(2θ)+sin(2θ))/(1+cos(2θ)+sin(2θ))=tan(θ)

Solución

verificar

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

Verdadero

Pasos de solución

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

Manipular el lado derecho

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

Re-escribir usando identidades trigonométricas

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

Multiplicar

\frac{1 + cos ( 2 θ )}{1 + cos ( 2 θ )}

por

Eq u a t i o n 1

= \frac{\frac{( 1 - c o s ( 2 θ ) ) ( 1 + c o s ( 2 θ ) )}{1 + c o s ( 2 θ )} + sin ( 2 θ )}{1 + cos ( 2 θ ) + sin ( 2 θ )}

Aplicar la siguiente regla para binomios al cuadrado:

2 θ^{2} - y^{2} = (2 θ + y) (2 θ - y)

(1 - cos (2 θ)) (1 + cos (2 θ)) = 1 - cos^{2} (2 θ)

= \frac{\frac{1 - c o s ^{2} ( 2 θ )}{1 + c o s ( 2 θ )} + sin ( 2 θ )}{1 + cos ( 2 θ ) + sin ( 2 θ )}

Utilizar la identidad pitagórica:

1 = cos^{2} (2 θ) + sin^{2} (2 θ)

1 - cos^{2} (2 θ) = sin^{2} (2 θ)

= \frac{\frac{s i n ^{2} ( 2 θ )}{1 + c o s ( 2 θ )} + sin ( 2 θ )}{1 + cos ( 2 θ ) + sin ( 2 θ )}

Simplificar

\frac{\frac{s i n ^{2} ( 2 θ )}{1 + c o s ( 2 θ )} + sin ( 2 θ )}{1 + cos ( 2 θ ) + sin ( 2 θ )} : \frac{sin ( 2 θ )}{1 + cos ( 2 θ )}

\frac{\frac{s i n ^{2} ( 2 θ )}{1 + c o s ( 2 θ )} + sin ( 2 θ )}{1 + cos ( 2 θ ) + sin ( 2 θ )}

Simplificar

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

en una fracción:

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

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

Convertir a fracción:

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

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

Ya que los denominadores son iguales, combinar las fracciones:

\frac{a}{c} \pm \frac{b}{c} = \frac{a \pm b}{c}

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

= \frac{\frac{s i n ^{2} ( 2 θ ) + s i n ( 2 θ ) ( c o s ( 2 θ ) + 1 )}{1 + c o s ( 2 θ )}}{1 + cos ( 2 θ ) + sin ( 2 θ )}

Aplicar las propiedades de las fracciones:

\frac{\frac{b}{c}}{a} = \frac{b}{c \cdot a}

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

Factorizar

sin^{2} (2 θ) + sin (2 θ) (1 + cos (2 θ)) : sin (2 θ) (sin (2 θ) + 1 + cos (2 θ))

sin^{2} (2 θ) + sin (2 θ) (1 + cos (2 θ))

Aplicar las leyes de los exponentes:

a^{b + c} = a^{b} a^{c}

sin^{2} (2 θ) = sin (2 θ) sin (2 θ)

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

Factorizar el termino común

sin (2 θ)

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

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

Eliminar los terminos comunes:

sin (2 θ) + 1 + cos (2 θ)

= \frac{sin ( 2 θ )}{1 + cos ( 2 θ )}

= \frac{sin ( 2 θ )}{1 + cos ( 2 θ )}

Utilizar la identidad trigonométrica del ángulo doble:

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

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

Utilizar la identidad trigonométrica del ángulo doble:

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

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

Simplificar

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

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

1 + 2 cos^{2} (θ) - 1 = 2 cos^{2} (θ)

1 + 2 cos^{2} (θ) - 1

Agrupar términos semejantes

= 2 cos^{2} (θ) + 1 - 1

1 - 1 = 0

= 2 cos^{2} (θ)

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

Dividir:

\frac{2}{2} = 1

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

Eliminar los terminos comunes:

cos (θ)

= \frac{sin ( θ )}{cos ( θ )}

= \frac{sin ( θ )}{cos ( θ )}

= \frac{sin ( θ )}{cos ( θ )}

Utilizar la identidad trigonométrica básica:

\frac{sin ( x )}{cos ( x )} = tan (x)

= tan (θ)

Se demostró que ambos lados pueden tomar la misma forma

\Rightarrow Verdadero

cot (x) sec^{2} (x) - cot (x) = tan (x)

\frac{cos ( θ )}{cos ( θ ) - 1} = \frac{1}{1 - sec ( θ )}

cos (- \frac{π}{4} - 5 t) = cos (5 t + \frac{π}{4})

tan (\frac{π}{4} - θ) = \frac{1 - tan ( θ )}{1 + tan ( θ )}

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