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verificar sin(60)=2sin(30)cos(30)

Solución

verificar

sin (6 0^{\circ}) = 2 sin (3 0^{\circ}) cos (3 0^{\circ})

Verdadero

Pasos de solución

sin (6 0^{\circ}) = 2 sin (3 0^{\circ}) cos (3 0^{\circ})

Manipular el lado derecho

sin (6 0^{\circ})

Simplificar

= \frac{3}{2}

Manipular el lado izquierdo

2 sin (3 0^{\circ}) cos (3 0^{\circ})

Simplificar

2 cos (3 0^{\circ}) sin (3 0^{\circ}) : \frac{3}{2}

2 cos (3 0^{\circ}) sin (3 0^{\circ})

Simplificar

cos (3 0^{\circ}) : \frac{3}{2}

cos (3 0^{\circ})

Utilizar la siguiente identidad trivial:

cos (3 0^{\circ}) = \frac{3}{2}

cos (x)

tabla de valores periódicos con

36 0^{\circ} n

intervalos:

x 0 3 0^{\circ} 4 5^{\circ} 6 0^{\circ} 9 0^{\circ} 12 0^{\circ} 13 5^{\circ} 15 0^{\circ} cos (x) 1 \frac{3}{2} \frac{2}{2} \frac{1}{2} 0 - \frac{1}{2} - \frac{2}{2} - \frac{3}{2} x 18 0^{\circ} 21 0^{\circ} 22 5^{\circ} 24 0^{\circ} 27 0^{\circ} 30 0^{\circ} 31 5^{\circ} 33 0^{\circ} cos (x) - 1 - \frac{3}{2} - \frac{2}{2} - \frac{1}{2} 0 \frac{1}{2} \frac{2}{2} \frac{3}{2}

= \frac{3}{2}

= 2 \cdot \frac{3}{2} sin (3 0^{\circ})

Simplificar

sin (3 0^{\circ}) : \frac{1}{2}

sin (3 0^{\circ})

Utilizar la siguiente identidad trivial:

sin (3 0^{\circ}) = \frac{1}{2}

tabla de valores periódicos con

36 0^{\circ} n

intervalos:

x 0 3 0^{\circ} 4 5^{\circ} 6 0^{\circ} 9 0^{\circ} 12 0^{\circ} 13 5^{\circ} 15 0^{\circ} sin (x) 0 \frac{1}{2} \frac{2}{2} \frac{3}{2} 1 \frac{3}{2} \frac{2}{2} \frac{1}{2} x 18 0^{\circ} 21 0^{\circ} 22 5^{\circ} 24 0^{\circ} 27 0^{\circ} 30 0^{\circ} 31 5^{\circ} 33 0^{\circ} sin (x) 0 - \frac{1}{2} - \frac{2}{2} - \frac{3}{2} - 1 - \frac{3}{2} - \frac{2}{2} - \frac{1}{2}

= \frac{1}{2}

= 2 \cdot \frac{1}{2} \cdot \frac{3}{2}

Simplificar

2 \cdot \frac{3}{2} \cdot \frac{1}{2}

Multiplicar fracciones:

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

= \frac{3 \cdot 2}{2 \cdot 2}

Eliminar los terminos comunes:

2

= \frac{3}{2}

= \frac{3}{2}

= \frac{3}{2}

Se demostró que ambos lados pueden tomar la misma forma

\Rightarrow Verdadero

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

cos (x) \cdot tan (x) = sin (x)

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

\frac{sec ( x )}{csc ( x )} + \frac{sin ( x )}{cos ( x )} = 2 tan (x)

\frac{sin ( θ ) + tan ( θ )}{1 + sec ( θ )} = sin (θ)